Meiosis Cell Division
Meiosis is reductional division, which takes place in reproductive cell. It is divided into meiosis I and meiosis II. Meiosis I is divided into prophase I, metaphase I, anaphase I, telophase I and mitosis into prophase II, metaphase II, anaphase II, and telophase II. This note has detail information on different stages of meiosis cell division.
Summary
Meiosis is reductional division, which takes place in reproductive cell. It is divided into meiosis I and meiosis II. Meiosis I is divided into prophase I, metaphase I, anaphase I, telophase I and mitosis into prophase II, metaphase II, anaphase II, and telophase II. This note has detail information on different stages of meiosis cell division.
Things to Remember
- Meiosis is reductional division, which takes place in reproductive cell. It is divided into meiosis I and meiosis II.
- Meiosis I is divided into prophase I, metaphase I, anaphase I, telophase I and mitosis into prophase II, metaphase II, anaphase II, and telophase II.
- Prophase is further divided leptotene, zygotene, pachytene, diplotene, and diakinasis.
- Meiosis creates variation in offspring and helps in reproduction and genetic stability.
MCQs
No MCQs found.
Subjective Questions
Q1:
Find the centre and radius of the circle whose equation is given by 2x2 + 2y2 - 5x - 7y - 23 = 0.
Type: Short Difficulty: Easy
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Answer: <p>Here,</p> <p>2x<sup>2</sup> + 2y<sup>2</sup> - 5x - 7y - 23 = 0</p> <p>or, \(\frac {2x^2}2\) + \(\frac {2y^2}2\) - \(\frac 52\)x - \(\frac 72\)y - \(\frac {23}2\) = 0</p> <p>or, x<sup>2</sup> - \(\frac 52\)x + y<sup>2</sup> - \(\frac 72\)y - \(\frac {23}2\) = 0</p> <p>or, x<sup>2</sup> - 2.x.\(\frac 54\) + (\(\frac 54)\)<sup>2</sup> + y<sup>2</sup> - 2.y.\(\frac 74\) + (\(\frac 74\))<sup>2</sup> + (\(\frac 74\))<sup>2</sup>- (\(\frac 74\))<sup>2</sup> - \(\frac {23}2\) = 0</p> <p>or, (x - \(\frac 54\))<sup>2</sup> + (y - \(\frac 74\))<sup>2</sup> = \(\frac {25}{16}\) + \(\frac {49}{16}\) + \(\frac {23}2\) = \(\frac {25 + 49 + 184}{16}\)</p> <p>or, (x - \(\frac 54\))<sup>2</sup> + (y - \(\frac 74\))<sup>2</sup> = \(\frac {258}{16}\)............................(1)</p> <p>Eq<sup>n</sup> of circle is: (x - h)<sup>2</sup>+ (y - x)<sup>2</sup> = r<sup>2</sup>......................(2)</p> <p>Comparing (1) and (2)</p> <p>Centre of circle (h, k) = (\(\frac 54\), \(\frac 74\))</p> <p>and radius of circle (r) = \(\frac {\sqrt {258}}4\) units <sub>Ans</sub></p>
Q2:
Find the co-ordinates of the centre of the circle:
x2 + y2 - 20y + 75 = 0
Type: Short Difficulty: Easy
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Answer: <p>Here,</p> <p>x<sup>2</sup> + y<sup>2</sup> - 20y + 75 = 0</p> <p>or, x<sup>2</sup> + y<sup>2</sup> - 2.y.10 + (10)<sup>2</sup> - (10)<sup>2</sup> + 75 = 0</p> <p>or, (x - 0)<sup>2</sup> + (y - 10)<sup>2</sup>- 100 + 75 = 0</p> <p>or, (x - 0)<sup>2</sup> + (y - 10)<sup>2</sup>- 25 = 0</p> <p>or, (x - 0)<sup>2</sup> + (y - 10)<sup>2</sup>= 25</p> <p>or, (x - 0)<sup>2</sup> + (y - 10)<sup>2</sup>=(5)<sup>2</sup>............................(1)</p> <p>The eq<sup>n</sup> of circle is: (x - h)<sup>2</sup> + (y - k)<sup>2</sup> = r<sup>2</sup>.....................(2)</p> <p>Comparing (1) and (2)</p> <p>Centre of circle (h, k) = (0. 10) <sub>Ans</sub></p>
Q3:
What will be the length of the radius of the circle whose equation is: x2 + y2 - 4x + 10y - 7 = 0?
Type: Short Difficulty: Easy
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Answer: <p>Here,</p> <p>x<sup>2</sup> + y<sup>2</sup> - 4x + 10y - 7 = 0</p> <p>or, x<sup>2</sup>- 4x + y<sup>2</sup> + 10y - 7 = 0</p> <p>or, x<sup>2</sup> - 2.x.2 + 2<sup>2</sup> - 2<sup>2</sup> + y<sup>2</sup> + 2.y.5 + 5<sup>2</sup> - 5<sup>2</sup> - 7 = 0</p> <p>or, (x - 2)<sup>2</sup> + (y + 5)<sup>2</sup> - 4 - 25 - 7 = 0</p> <p>or, (x - 2)<sup>2</sup> + (y + 5)<sup>2</sup> - 36 = 0</p> <p>or, (x - 2)<sup>2</sup> + (y + 5)<sup>2</sup>= 36</p> <p>or, (x - 2)<sup>2</sup> + (y + 5)<sup>2</sup>= 6<sup>2</sup>...........................(1)</p> <p>The eq<sup>n</sup> of the circle is: (x - h)<sup>2</sup>+ (y - k)<sup>2</sup> = r<sup>2</sup>.........................(2)</p> <p>Comparing (1) and (2)</p> <p>The length of radius of a circle is: 6 units <sub>Ans</sub></p> <p></p> <p></p>
Q4:
Calculate the length of the radius of the circle whose equation is: x2 + y2 + 4x - 6y + 4 = 0.
Type: Short Difficulty: Easy
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Answer: <p>Here,</p> <p>x<sup>2</sup> + y<sup>2</sup> + 4x - 6y + 4 = 0</p> <p>or, x<sup>2</sup> + 4x+ 4 + y<sup>2</sup> - 6y = 0</p> <p>or, x<sup>2</sup> + 2.x.2 + 2<sup>2</sup>+ y<sup>2</sup> - 2.y.3 + 3<sup>2</sup> - 3<sup>2</sup>= 0</p> <p>or, (x + 2)<sup>2 </sup>+ (y - 3)<sup>2</sup> = 3<sup>2</sup>.....................................(1)</p> <p>The eq<sup>n</sup> of circle is: (x - h)<sup>2</sup> + (y - k)<sup>2</sup> = r<sup>2</sup>...........................(2)</p> <p>Comparing (1) and (2)</p> <p>The length of radius of the circle (r) = 3 units<sub>Ans</sub></p>
Q5:
Find the equation of a circle having centre (3, 0) and radius 5 units.
Type: Short Difficulty: Easy
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Answer: <p>Here,</p> <p>Centre of circle (h, k) = (3, 0)</p> <p>Radius of circle (r) = 5 units</p> <p>The equation of circle is: (x - h)<sup>2 + </sup>(y - k)<sup>2</sup> = r<sup>2</sup></p> <p>(x - 3)<sup>2</sup> + (y - 0)<sup>2</sup> = 5<sup>2</sup></p> <p>or, x<sup>2</sup> - 6x + 9 + y<sup>2</sup> = 25</p> <p>or, x<sup>2</sup> + y<sup>2</sup> - 6x + 9 - 25 = 0</p> <p>∴x<sup>2</sup> + y<sup>2</sup> - 6x -16 = 0<sub>Ans</sub></p>
Q6:
Find the centre and radius of x2 + y2 + 4x - 4y -1 = 0.
Type: Short Difficulty: Easy
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Answer: <p>Here,</p> <p>x<sup>2</sup> + y<sup>2</sup> + 4x - 4y -1 = 0</p> <p>or, x<sup>2</sup> + 4x + y<sup>2</sup> - 4y - 1 = 0</p> <p>or, x<sup>2</sup> + 2.x.2 + 2<sup>2</sup> - 2<sup>2</sup> + y<sup>2</sup> - 2.y.2 + 2<sup>2</sup> - 2<sup>2</sup> - 1 = 0</p> <p>or, (x + 2)<sup>2</sup> + (y - 2)<sup>2</sup> - 9 = 0</p> <p>or, (x + 2)<sup>2</sup> + (y - 2)<sup>2</sup>= 9</p> <p>or, (x + 2)<sup>2</sup> + (y - 2)<sup>2</sup>=3<sup>2</sup>................................(1)</p> <p>Equation of circle, (x - h)<sup>2</sup> + (y - k)<sup>2</sup> = r<sup>2</sup>...................................(2)</p> <p>Comparing (1) and (2)</p> <p>h = -2</p> <p>k = 2</p> <p>r = 3</p> <p>∴ Centre = (-2, 2) and radius = 3 units<sub>Ans</sub></p>
Q7:
If the equation of the circle is: x2 + y2 - 2y = 24, find its radius and its centre.
Type: Short Difficulty: Easy
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Answer: <p>Here,</p> <p>x<sup>2</sup>+ y<sup>2</sup> - 2y = 24</p> <p>or, x<sup>2</sup> + y<sup>2</sup> - 2.y.1 + 1<sup>2</sup>- 1<sup>2</sup> = 24</p> <p>or, x<sup>2</sup> + (y - 1)<sup>2</sup> - 1 = 24</p> <p>or, x<sup>2</sup> + (y - 1)<sup>2</sup> = 25</p> <p>or, x<sup>2</sup> + (y - 1)<sup>2</sup> = 5<sup>2</sup>.............................(1)</p> <p>The equation of circle is: (x - h)<sup>2</sup> + (y - k)<sup>2</sup> = a<sup>2</sup>........................(2)</p> <p>Comparing (1) and (2)</p> <p>(h, k) = (0, 1)</p> <p>a = 5</p> <p>Hence, centre = (0, 1) and radius = 5 units<sub>Ans</sub></p>
Q8:
If the equation of a circle is (x + 5)2 + y2 = 121, find the co-ordinates of the centre of the circle and its diameter.
Type: Short Difficulty: Easy
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Answer: <p>Given eq<sup>n</sup> is:</p> <p>(x + 5)<sup>2</sup> + y<sup>2</sup> = 121</p> <p>or, (x + 5)<sup>2</sup> + (y - 0)<sup>2</sup> = (11)<sup>2</sup>..........................(1)</p> <p>The eq<sup>n</sup> of the circle is:</p> <p>(x - h)<sup>2</sup> + (y - k)<sup>2</sup> = r<sup>2</sup>..............................(2)</p> <p>Comparing (1) and (2)</p> <p>h = -5</p> <p>k = 0</p> <p>r = 11</p> <p>∴ The centre of circle (h, k) = (-5, 0) and radius (r) = 11</p> <p>∴ Diameter = 2r = 2× 11 = 22 units<sub>Ans</sub></p>
Q9:
The equation of a circle is: x2 + y2 - 4x - 6y - 12 = 0. Find the co-ordinates of its centre.
Type: Short Difficulty: Easy
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Answer: <p>Here,</p> <p>x<sup>2</sup> + y<sup>2</sup> - 4x - 6y - 12 = 0</p> <p>or, x<sup>2</sup> - 4x + y<sup>2</sup> - 6y - 12 = 0</p> <p>or, x<sup>2</sup> - 2.x.2 + 2<sup>2</sup> - 2<sup>2</sup> + y<sup>2</sup>- 2.y.3 + 3<sup>2</sup>- 3<sup>2</sup>- 12 = 0</p> <p>or, (x - 2)<sup>2</sup> + (y - 3)<sup>2</sup> - 12 - 4 - 9= 0</p> <p>or, (x - 2)<sup>2</sup> + (y - 3)<sup>2</sup> - 25 = 0</p> <p>or, (x - 2)<sup>2</sup> + (y - 3)<sup>2</sup> = 25</p> <p>or, (x - 2)<sup>2</sup> + (y - 3)<sup>2</sup> = 5<sup>2</sup>............................(1)</p> <p>The equation of circle is: (x - h)<sup>2</sup> + (y - k)<sup>2</sup> = r<sup>2</sup>.............................(2)</p> <p>Comparing (1) and (2)</p> <p>(h, k) = (3, 2) and r = 5</p> <p>∴ The centre of circle = (3, 2) and radius of circle (r) = 5 units<sub>Ans</sub></p>
Q10:
Find the equation of a circle whose ends of diameter has co-ordinates (2, 4) and (3, -6).
Type: Short Difficulty: Easy
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Answer: <p>Here,</p> <p>(x<sub>1</sub> - y<sub>1</sub>) = (2, 4)</p> <p>(x<sub>2</sub> - y<sub>2</sub>) = (3, -6)</p> <p>The equation of a circle in diameter form:</p> <p>(x - x<sub>1</sub>) (x - x<sub>2</sub>) + (y - y<sub>1</sub>) (y - y<sub>2</sub>) = 0</p> <p>or, (x - 2) (x - 3) + (y - 4) (y + 6) = 0</p> <p>or, x<sup>2</sup> - 3x - 2x + 6 + y<sup>2</sup> + 6y - 4y - 24 = 0</p> <p>or, x<sup>2</sup> - 5x + 6 + y<sup>2</sup> + 2y - 24 = 0</p> <p>∴ x<sup>2</sup> + y<sup>2</sup> - 5x + 2y - 18 = 0<sub>Ans</sub></p>
Q11:
Find the equation of the given circle.
Type: Short Difficulty: Easy
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Answer: <p></p> <figure class="inline-right" style="width: 150px;"><img src="/uploads/j2.jpg" alt="z" width="150" height="150"><figcaption><br></figcaption></figure><p>Let: A(4, 2) and B(3, 5) are the two ends of the diameter of a circle.</p> <p>The equation of a circle in diameter form:</p> <p>(x - x<sub>1</sub>) (x - x<sub>2</sub>) + (y - y<sub>1</sub>) (y - y<sub>2</sub>) = 0</p> <p>or, (x, 4) (x, 3) + (y - 2) (y - 5) = 0</p> <p>or, x<sup>2</sup> - 3x - 4x + 12 + y<sup>2</sup> - 5y - 2y + 10 = 0</p> <p>∴ x<sup>2</sup> + y<sup>2</sup> - 7x - 7y + 22 = 0<sub>Ans</sub></p>
Q12:
Find the equation of the circle of its radius is 5 units and equations of two diameters are: x = 3y and y = 2.
Type: Short Difficulty: Easy
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Answer: <p>Here,</p> <p>Radius of circle (r) = 5 units</p> <p>Eq<sup>n</sup> of two diameters are:</p> <p>x = 3y.............................(1)</p> <p>y = 2...............................(2)</p> <p>Putting the value of y in eq<sup>n</sup> (1)</p> <p>x = 3× 2 = 6</p> <p>Intersection point of the two diameters is centre of circle so:</p> <p>Centre of circle = (h, k) = (6, 2)</p> <p>Eq<sup>n</sup> of circle is:</p> <p>(x - h)<sup>2</sup> + (y - k)<sup>2</sup> = r<sup>2</sup></p> <p>or, (x - 6)<sup>2</sup> + (y - 2)<sup>2</sup> = 5<sup>2</sup></p> <p>or, x<sup>2</sup>- 12x + 36 + y<sup>2</sup> - 4y + 4 - 25 = 0</p> <p>∴ x<sup>2</sup> + y<sup>2</sup> - 12x - 4y + 15 = 0<sub>Ans</sub></p>
Q13:
Find the equation of the circle with centre (-2, 3) and touching on the x-axis.
Type: Short Difficulty: Easy
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Answer: <p>Here,</p> <p>Centre of circle (h, k) = (-2, 3)</p> <p>Radius (r) = k = 3</p> <p>The eq<sup>n</sup> of circle is:</p> <p>(x - h)<sup>2</sup> + (y - k)<sup>2</sup> = r<sup>2</sup></p> <p>or, (x + 2)<sup>2</sup> + (y - 3)<sup>2</sup> = 3<sup>2</sup></p> <p>or, x<sup>2</sup> + 4x + 4 + y<sup>2</sup> - 6y + 9 = 9</p> <p>or, x<sup>2</sup> + y<sup>2</sup>+ 4x - 6y + 13 - 9 = 0</p> <p>∴ x<sup>2</sup> + y<sup>2</sup>+ 4x - 6y + 4 = 0<sub>Ans</sub></p>
Q14:
Find the equation of the circle with centre (4, -3) and touching on y-axis.
Type: Short Difficulty: Easy
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Answer: <p>Here,</p> <p>Centre of circle (h, k) = (4, -3)</p> <p>Radius (r) = h = 4</p> <p>The equation of circle is:</p> <p>(x - h)<sup>2</sup> + (y - k)<sup>2</sup> = r<sup>2</sup></p> <p>or, (x - 4)<sup>2</sup> + (y + 3)<sup>2</sup> = 4<sup>2</sup></p> <p>or, x<sup>2</sup> - 8x + 16 + y<sup>2</sup> + 6y + 9 = 16</p> <p>or, x<sup>2</sup> + y<sup>2</sup> - 8x + 6y + 25 = 16</p> <p>or, x<sup>2</sup> + y<sup>2</sup> - 8x + 6y + 25 -16 = 0</p> <p>∴ x<sup>2</sup> + y<sup>2</sup> - 8x + 6y + 9 = 0<sub>Ans</sub></p>
Q15:
Find equation of the circle which touches the positive x-axis and y-axis, whose radius is 5 units.
Type: Short Difficulty: Easy
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Answer: <p>Here,</p> <p>Radius of circle (r) = 5 units</p> <p>h = k = r = 5 [\(\because\) touches on both axis]</p> <p>The eq<sup>n</sup> of circle is:</p> <p>(x - h)<sup>2</sup> + (y - k)<sup>2</sup> = r<sup>2</sup></p> <p>or, (x - 5)<sup>2</sup> + (y - 5)<sup>2</sup> = 5<sup>2</sup></p> <p>or, x<sup>2</sup> - 10x + 25 + y<sup>2</sup> - 10y + 25 = 25</p> <p>or, x<sup>2</sup> + y<sup>2</sup> - 10x - 10y + 50 - 25 = 0</p> <p>∴ x<sup>2</sup> + y<sup>2</sup> - 10x - 10y + 25 = 0<sub>Ans</sub></p>
Q16:
Find the equation of the circle with centre of (6, 8) and passing through the origin.
Type: Short Difficulty: Easy
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Answer: <p>Here,</p> <p>Centre of circle = (h, k) = (6, 8)</p> <p>The eq<sup>n</sup> of circle is:</p> <p>(x - h)<sup>2</sup> + (y - k)<sup>2</sup> = r<sup>2</sup></p> <p>or, (x - 6)<sup>2</sup> + (y - 8)<sup>2</sup> = r<sup>2</sup>..............................(1)</p> <p>The eq<sup>n</sup> (1) passes through the point (0, 0)</p> <p>(0 - 6)<sup>2</sup>+ (0 - 8)<sup>2</sup> = r<sup>2</sup></p> <p>or, 36 + 64 = r<sup>2</sup></p> <p>or, 100 = r<sup>2</sup></p> <p>or, (r)<sup>2</sup> = (10)<sup>2</sup></p> <p>∴ r = 10 units</p> <p>Putting the value of r in eq<sup>n</sup> (1)</p> <p>(x - 6)<sup>2</sup> + (y - 8)<sup>2</sup> = (10)<sup>2</sup></p> <p>or, x<sup>2</sup>- 12x + 36 + y<sup>2</sup> - 16y + 64 = 100</p> <p>or, x<sup>2</sup> + y<sup>2</sup> - 12x - 16y + 100 = 100</p> <p>∴x<sup>2</sup> + y<sup>2</sup> - 12x - 16y= 0<sub>Ans</sub></p>
Q17:
To find the equation of the circle, whose ends of the diameter are at the points A(x1, y1) and B(x2, y2).
Type: Long Difficulty: Easy
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Answer: <p></p> <figure class="inline-right" style="width: 200px;"><img src="/uploads/k2.jpg" alt="z" width="200" height="149"><figcaption><br></figcaption></figure><p>Here,</p> <p>AB is the diameter whose co-ordinates areA(x<sub>1</sub>, y<sub>1</sub>) and B(x<sub>2</sub>, y<sub>2</sub>).</p> <p>Let: P(x, y) be any point on the circle.</p> <p>Slope of PA = \(\frac {y - y_1}{x - x_1}\)</p> <p>Slope of PB = \(\frac {y - y_2}{x - x_2}\)</p> <p>\(\angle\)APB = 90° [\(\because\) The angle made in a semi-circle is a right angle.]</p> <p>Now,</p> <p>slope of PA× slope of PB = -1</p> <p>or,\(\frac {y - y_1}{x - x_1}\)× \(\frac {y - y_2}{x - x_2}\) = -1</p> <p>or, (y - y<sub>1</sub>) (y - y<sub>2</sub>) = - (x - x<sub>1</sub>) (x - x<sub>2</sub>)</p> <p>∴(x - x<sub>1</sub>) (x - x<sub>2</sub>) + (y - y<sub>1</sub>) (y - y<sub>2</sub>) = 0<sub>Ans</sub></p>
Q18:
Find the centre and radius of the circle passing through the points P(2, -2), Q(6, 6) and R(5, 7). Also find the equation of the circle.
Type: Long Difficulty: Easy
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Answer: <p>Here,</p> <p>Given points of the circle are:P(2, -2), Q(6, 6) and R(5, 7)</p> <p>Let, centre of circle is: (h, k) and radius of circle is: a.</p> <p>Equation of circle is (x - h)<sup>2</sup> + (y - k)<sup>2</sup> = a<sup>2</sup>...........................(1)</p> <p>Given points P,Q and R passes through the equation (1)</p> <p>(2 - h)<sup>2</sup> + (-2, - k)<sup>2</sup>= a<sup>2</sup>.........................(2)</p> <p>(6 - h)<sup>2</sup> + (6 - k)<sup>2</sup> = a<sup>2</sup>...........................(3)</p> <p>(5 - h)<sup>2</sup> + (7 - k)<sup>2</sup> = a<sup>2</sup>...........................(4)</p> <p>From equation (2) and (3)</p> <p>(2 - h)<sup>2</sup> + (-2, - k)<sup>2</sup>=(6 - h)<sup>2</sup> + (6 - k)<sup>2</sup></p> <p>or, 4 - 4h + h<sup>2</sup> + 4 + 4k + k<sup>2</sup> = 36 - 12h + h<sup>2</sup> + 36 - 12k + k<sup>2</sup></p> <p>or, 8h + 16k + 8 = 72</p> <p>or, 8h + 16k = 72 - 8</p> <p>or, 8(h + 2k) = 64</p> <p>or, h + 2k = 8.......................................(5)</p> <p>Similarly,</p> <p>From equation (2) and (4)</p> <p>(2 - h)<sup>2</sup> + (-2, - k)<sup>2</sup>=(5 - h)<sup>2</sup> + (7 - k)<sup>2</sup></p> <p>or,4 - 4h + h<sup>2</sup> + 4 + 4k + k<sup>2</sup>= 25 - 10h + h<sup>2</sup> + 49 - 14k + k<sup>2</sup></p> <p>or, -4h + 4k + 8 = 74 - 10h - 14k</p> <p>or, -4h + 10h + 4k + 14k = 74 - 8</p> <p>or, 6h + 18k = 66</p> <p>or, 6(h + 3k) = 66</p> <p>or, h + 3k = 11......................................(6)</p> <p>Subtracting equation (6) from equation (5)</p> <table width="134"><tbody><tr><td>h</td> <td>+</td> <td>2k</td> <td>=</td> <td>8</td> </tr><tr><td>h</td> <td>+</td> <td>3k</td> <td>=</td> <td>11</td> </tr><tr><td>-</td> <td>-</td> <td></td> <td></td> <td>-</td> </tr><tr><td></td> <td>-</td> <td>k</td> <td>=</td> <td>-3</td> </tr></tbody></table><p>∴ k = 3</p> <p>Substituting the value of k in equation (6)</p> <p>h + 3k = 11</p> <p>or, h + 3× 3 = 11</p> <p>or, h + 9 = 11</p> <p>or, h = 11 - 9</p> <p>∴ h = 2</p> <p>∴ Centre of circle (h, k) = (2, 3)</p> <p>Again,</p> <p>Substituting the value of h, k in equation (2)</p> <p>(2 - 2)<sup>2</sup> + (-2 - 3)<sup>2</sup> = a<sup>2</sup></p> <p>or, 0 + (-5)<sup>2</sup> = a<sup>2</sup></p> <p>or, a<sup>2</sup> = 25</p> <p>∴ a = 5</p> <p>Substituting the value of (h, k) and 'a' in equation (1)</p> <p>(x - 2)<sup>2</sup> + (y - 3)<sup>2</sup> = 5<sup>2</sup></p> <p>or, x<sup>2</sup> - 4x + 4 + y<sup>2</sup> - 6y + 9 = 25</p> <p>or, x<sup>2</sup> + y<sup>2</sup> - 4x - 6y + 13 - 25 = 0</p> <p>∴x<sup>2</sup> + y<sup>2</sup> - 4x - 6y -12 = 0<sub>Ans</sub></p>
Q19:
Find the equation of the circle which passes through the points (4, 1) and (6, 5) and whose centre lies on the line 4x + y = 16.
Type: Long Difficulty: Easy
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Answer: <p>Here,</p> <p>Centre of circle (h, k) and radius of circle = a.</p> <p>Equation of the circle is: (x - h)<sup>2</sup> + (y - k)<sup>2</sup> = a<sup>2</sup>........................(1)</p> <p>The points (4, 1) and (6, 5) passes through the equation (1)</p> <p>(4 - h)<sup>2</sup> + (1 - k)<sup>2</sup> = a<sup>2</sup>.....................(2)</p> <p>(6 - h)<sup>2</sup> + (5 - k)<sup>2</sup> = a<sup>2</sup>.....................(3)</p> <p>From equation (2) and (3)</p> <p>(6 - h)<sup>2</sup> + (5 - k)<sup>2</sup>=(4 - h)<sup>2</sup> + (1 - k)<sup>2</sup></p> <p>or, 36 - 12h + h<sup>2</sup> + 25 - 10k + k<sup>2</sup> = 16 - 8h + h<sup>2</sup> + 1 - 2k + k<sup>2</sup></p> <p>or, 4h + 8k = 44</p> <p>or, 4(h + 2k) = 44</p> <p>or, h + 2k = 11..........................(4)</p> <p>Now,</p> <p>Centre (h, k) lies on the line 4x + y = 16</p> <p>4h + k = 16</p> <p>k = 16 - 4h...........................(5)</p> <p>Substituting the value of k in equation (4)</p> <p>h + 2(16 - 4h) = 11</p> <p>or, h + 32 - 8h = 11</p> <p>or, -7h = 11 - 32</p> <p>or, -7h = -21</p> <p>or, h = \(\frac {-21}{-7}\)</p> <p>∴ h = 3</p> <p>Substituting the value of h in equation (5)</p> <p>∴ k = 16 - 4× 3 = 16 - 12 = 4</p> <p>Substituting (h, k) in equation (2)</p> <p>(4 - 3)<sup>2</sup> + (1 - 4)<sup>2</sup> = a<sup>2</sup></p> <p>or, a<sup>2</sup>= 1<sup>2</sup>+ (-3)<sup>2</sup></p> <p>or, a<sup>2</sup> = 1 + 9</p> <p>or, a<sup>2</sup> = 10</p> <p>∴ a = \(\sqrt {10}\)</p> <p>The equation of circle is:</p> <p>(x - h)<sup>2</sup> + (y - k)<sup>2</sup> = a<sup>2</sup></p> <p>or, (x - 3)<sup>2</sup> + (y - 4)<sup>2</sup> = (\(\sqrt {10}\))<sup>2</sup></p> <p>or, x<sup>2</sup> - 2.x.3 + 3<sup>2</sup> + y<sup>2</sup> - 2.y.4 + 4<sup>2</sup> = 10</p> <p>or, x<sup>2</sup> + y<sup>2</sup> - 6x - 8y + 9 + 16 = 10</p> <p>or, x<sup>2</sup> + y<sup>2</sup> - 6x + 8y + 25 - 10 = 0</p> <p>∴x<sup>2</sup> + y<sup>2</sup> - 6x + 8y + 15 = 0<sub>Ans</sub></p>
Q20:
If one end of a diameter of a circle x2 + y2 - 4x - 6y + 11 = 0 is (3, 4), find the co-ordinates of the other end.
Type: Long Difficulty: Easy
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Answer: <p>Here,</p> <p>x<sup>2</sup> + y<sup>2</sup> - 4x - 6y + 11 = 0</p> <p>or, x<sup>2</sup> - 4x + y<sup>2</sup> - 6y + 11 = 0</p> <p>or, x<sup>2</sup> - 2.x.2 + 2<sup>2</sup> - 2<sup>2</sup> + y<sup>2</sup> - 2.y.3 + 3<sup>2</sup> - 3<sup>2</sup> + 11 = 0</p> <p>or, (x - 2)<sup>2</sup> + (y - 3)<sup>2</sup> - 4 - 9 + 11 = 0</p> <p>or, (x - 2)<sup>2</sup> + (y - 3)<sup>2</sup> - 2= 0</p> <p>or, (x - 2)<sup>2</sup> + (y - 3)<sup>2</sup>=2....................................(1)</p> <p>The equation of circle is:</p> <p>(x - h)<sup>2</sup> + (y - k)<sup>2</sup> = r<sup>2</sup>..............................(2)</p> <p>Comparing (1) and (2)</p> <p>(h, k) = (2, 3)</p> <p>∴ Centre of circle is a mid-point of the diameter of the circle is (2, 3)</p> <p>Mid-point of circle = (\(\frac {x_1 + x_2}2\), \(\frac {y_1 + y_2}2\))</p> <p>or, (2, 3) = (\(\frac {3 + x}2\), \(\frac {4 + y}2\))</p> <table width="315"><tbody><tr><td> <p>2 = \(\frac {3 + x}2\)</p> <p>or, 3 + x = 4</p> <p>or, x = 4 - 3</p> <p>∴ x = 1</p> </td> <td> <p>3 = \(\frac {4 + y}2\)</p> <p>or, 4 + y = 6</p> <p>or, y = 6 - 4</p> <p>∴ y = 2</p> </td> </tr></tbody></table><p>∴ The other point is: (1, 2)<sub>Ans</sub></p>
Q21:
Find the co-ordinates of center and length of diameter of the circle:
2x2 + 2y2 - 6x - 2y - 13 = 0
Type: Long Difficulty: Easy
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Answer: <p>Here,</p> <p>2x<sup>2</sup> + 2y<sup>2</sup> - 6x - 2y - 13 = 0</p> <p>or, x<sup>2</sup> + y<sup>2</sup> - 3x - y - \(\frac {13}2\) = 0 [\(\because\) Dividing by 2 on both sides]</p> <p>or, x<sup>2</sup> - 3x + y<sup>2</sup> - y - \(\frac {13}2\)</p> <p>or, x<sup>2</sup> - 2.x.\(\frac 32\) + (\(\frac 32\)<sup>2</sup> - (\(\frac 32\)<sup>2</sup>+ y<sup>2</sup>- 2.y.\(\frac 12\) + (\(\frac 12\))<sup>2</sup> - (\(\frac 12\))<sup>2</sup> - \(\frac {13}2\) = 0</p> <p>or, (x - \(\frac 32\))<sup>2</sup> + (y - \(\frac 12\))<sup>2</sup> = \(\frac 94\) + \(\frac 14\) + \(\frac {13}2\)</p> <p>or, (x - \(\frac 32\))<sup>2</sup> + (y - \(\frac 12\))<sup>2</sup> = \(\frac {9 + 1 + 26}4\)</p> <p>or, (x - \(\frac 32\))<sup>2</sup> + (y - \(\frac 12\))<sup>2</sup> = \(\frac {36}4\)</p> <p>or, (x - \(\frac 32\))<sup>2</sup> + (y - \(\frac 12\))<sup>2</sup> = (\(\frac 62\))<sup>2</sup>...............................(1)</p> <p>The equation of circle is:</p> <p>(x - h)<sup>2</sup> + (y - k)<sup>2</sup> = r<sup>2</sup>....................................(2)</p> <p>Comparing equation (1) and (2)</p> <p>(h, k) = (\(\frac 32\), \(\frac 12\))</p> <p>radius (a) = \(\frac 62\) units</p> <p>Length of the diameter = 2r = 2× \(\frac 62\) = 6 units</p> <p>∴ Centre =(\(\frac 32\), \(\frac 12\)) and diameter = 6 units<sub>Ans</sub></p>
Q22:
The equation of a circle is: x2 + y2 - 2x - 6y + 1 = 0, find the co-ordinates of its centre and radius.
Type: Long Difficulty: Easy
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Answer: <p>Given equation is:</p> <p>x<sup>2</sup> + y<sup>2</sup> - 2x - 6y + 1 = 0</p> <p>or, x<sup>2</sup> - 2x + 1 + y<sup>2</sup> - 6y = 0</p> <p>or, x<sup>2</sup> - 2.x.1 + 1<sup>2</sup> + y<sup>2</sup> - 2.y.3 + 3<sup>2</sup> - 3<sup>2</sup> = 0</p> <p>or, (x - 1)<sup>2</sup> + (y - 3)<sup>2</sup> - 9 = 0</p> <p>or, (x - 1)<sup>2</sup> + (y - 3)<sup>2</sup>=3<sup>2</sup>............................(1)</p> <p>The equation of circle is: (x - h)<sup>2</sup> + (y - k)<sup>2</sup> = r<sup>2</sup>....................................(2)</p> <p>Comparing equation (1) and (2)</p> <p>∴ Centre of circle (h, k) = (1, 3)</p> <p>and radius (r) = 3 units<sub>Ans</sub></p>
Q23:
The co-ordinates of an end point of the diameter of circle x2 + y2 - 2x - 2y = 8 is (2, 4), find the co-ordinates of other end.
Type: Long Difficulty: Easy
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Answer: <p>Given equation is:</p> <p>x<sup>2</sup> + y<sup>2</sup> - 2x - 2y = 8</p> <p>or, x<sup>2</sup>- 2.x.1 + 1<sup>2</sup> - 1<sup>2</sup> + y<sup>2</sup>- 2.y.1 + 1<sup>2</sup> - 1<sup>2</sup> = 8</p> <p>or, (x - 2)<sup>2</sup> + (y - 2)<sup>2</sup> - 1 - 1 = 8</p> <p>or, (x - 2)<sup>2</sup> + (y - 2)<sup>2</sup>= 8 + 1 + 1</p> <p>or, (x - 2)<sup>2</sup> + (y - 2)<sup>2</sup>= \(\sqrt {10}\)..........................(1)</p> <p>The eq<sup>n</sup>of circle is:</p> <p>(x - h)<sup>2</sup> + (y - k)<sup>2</sup>= r<sup>2</sup>.................................(2)</p> <p>Comparing equation (1) and (2)</p> <p>h = 1</p> <p>k = 1</p> <p>r = \(\sqrt {10}\) units</p> <p>(h, k) = (1, 1)</p> <p>From the above figure,</p> <p>mid-point of AB = (\(\frac {x_1 + x_2}2\), \(\frac {y_1 + y_2}2\))</p> <p>or, (1, 1) = (\(\frac {x_1 + 2}2\), \(\frac {y_1 + 4}2\))</p> <table width="364"><tbody><tr><td> <p>1 = \(\frac {x + 2}2\)</p> <p>or, x + 2 = 2</p> <p>or, x = 2 - 2</p> <p>∴ x = 0</p> </td> <td> <p>1 = \(\frac {y + 4}2\)</p> <p>or, y + 4 = 2</p> <p>or, y = 2 - 4</p> <p>∴ y = -2</p> </td> </tr></tbody></table><p>∴ The other end of the diameter is: (0, -2)<sub>Ans</sub></p>
Q24:
Find the co-ordinates of the centre and the radius of the circle given by the equation:
2x2 + 2y2 - 8x - 12y + 1 = 0
Type: Long Difficulty: Easy
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Answer: <p>Here,</p> <p>Given equation is:</p> <p>2x<sup>2</sup> + 2y<sup>2</sup> - 8x - 12y + 1 = 0</p> <p>or, \(\frac {2x^2}2\) + \(\frac {2y^2}2\) - \(\frac {8x}2\) - \(\frac {12y}2\) + \(\frac 12\) = 0 [\(\because\) dividing by 2 on both sides]</p> <p>or, x<sup>2</sup> + y<sup>2</sup> - 4x - 6y + \(\frac 12\) = 0</p> <p>or, x<sup>2</sup>- 4x + y<sup>2</sup>- 6y + \(\frac 12\) = 0</p> <p>or, x<sup>2</sup> - 2.x.2 + 2<sup>2</sup> - 2<sup>2</sup>+ y<sup>2</sup> - 2.y.3 + 3<sup>2</sup> - 3<sup>2</sup> + \(\frac 12\) = 0</p> <p>or, (x - 2)<sup>2</sup>+ (y - 3)<sup>2</sup>- 4 - 9 + \(\frac 12\) = 0</p> <p>or, (x - 2)<sup>2</sup> + (y - 3)<sup>2</sup> = 13 - \(\frac 12\)</p> <p>or, (x - 2)<sup>2</sup> + (y - 3)<sup>2</sup> =\(\frac {26 - 1}2\)</p> <p>or, (x - 2)<sup>2</sup> + (y - 3)<sup>2</sup> =\(\frac {25}2\)</p> <p>or, (x - 2)<sup>2</sup> + (y - 3)<sup>2</sup> =(\(\frac 5{\sqrt 2})\)<sup>2</sup>........................(1)</p> <p>The eq<sup>n</sup> of circle is: (x - h)<sup>2</sup> + (y - k)<sup>2</sup>= r<sup>2</sup>...........................(2)</p> <p>Comparing equation (1) and (2)</p> <p>∴Centre of circle = (h, k) = (2, 3)</p> <p>and Radius of circle (r) =(\(\frac 5{\sqrt 2})\)<sub>Ans</sub></p>
Q25:
Find the equation of a circle which passes through the point (-2, 0) and (0. -2) and its centre lies on the straight lines 2x - 3y + 1 = 0.
Type: Long Difficulty: Easy
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Answer: <p></p> <figure class="inline-right" style="width: 200px;"><img src="/uploads/l.jpg" alt="c" width="200" height="230"><figcaption><br></figcaption></figure><p>Let: centre of circle be (h. k).</p> <p>Distance of OA = Distance of OB</p> <p>or, \(\sqrt {(h - 0)^2 + (k + 2)^2}\) = \(\sqrt {(h + 2)^2 + (k - 0)^2}\)</p> <p>Squaring on both sides,</p> <p>h<sup>2</sup> + k<sup>2</sup> + 4k + 4 = h<sup>2</sup> + 4h + 4 + k<sup>2</sup></p> <p>or, 4k = 4h</p> <p>or, k - h = 0.........................(1)</p> <p>The centre (h, k) lies on the line:</p> <p>2x - 3y + 1 = 0</p> <p>2h - 3k + 1 = 0....................(2)</p> <p>Multiply eq<sup>n</sup> (1) by 2 and adding with eq<sup>n</sup> (2)</p> <table width="121"><tbody><tr><td>-2h</td> <td>+</td> <td>2k</td> <td>=</td> <td>0</td> </tr><tr><td>2h</td> <td>-</td> <td>3k</td> <td>=</td> <td>-1</td> </tr><tr><td></td> <td>-</td> <td>k</td> <td>=</td> <td>-1</td> </tr></tbody></table><p>∴ k = 1</p> <p>Putting the value of k in eq<sup>n</sup> (1)</p> <p>k - h = 0</p> <p>or, 1 - h = 0</p> <p>or, -h = -1</p> <p>∴ h = 1</p> <p>∴ Centre (h, k) = (1, 1)</p> <p>\begin{align*} Radius (r) &= \sqrt {(x - h)^2 + (y - k)^2}\\ &= \sqrt {(0 - 1)^2 + (-2 - 1)^2}\\ &= \sqrt {1 + 9}\\ &=\sqrt {10} units\\ \end{align*}</p> <p>Equation of circle is:</p> <p>(x - h)<sup>2</sup> + (y - k)<sup>2</sup> = r<sup>2</sup></p> <p>or, (x - 1)<sup>2</sup> + (y - 1)<sup>2</sup> = (\(\sqrt {10}\))<sup>2</sup></p> <p>or, x<sup>2</sup> - 2.x1 + 1<sup>2</sup> + y<sup>2</sup> - 2.y.1 + 1<sup>2</sup> = 10</p> <p>or, x<sup>2</sup> + y<sup>2</sup> - 2x - 2y + 1 + 1 - 10= 0</p> <p>∴x<sup>2</sup> + y<sup>2</sup> - 2x - 2y - 8 = 0<sub>Ans</sub></p> <p></p>
Q26:
The centre of the circle lies on the line 2x + y - 1 = 0. If it is passing through the point (4, 3) and its radius is 5 units. Find the equation of the circle.
Type: Long Difficulty: Easy
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Answer: <p></p> <figure class="inline-right" style="width: 200px;"><img src="/uploads/m2.jpg" alt="f" width="200" height="230"><figcaption><br></figcaption></figure><p>Let: the centre of circle = (h, k)</p> <p>The centre of a circle having radius 5 units (h, k) lies on the line 2x + y - 1 = 0.</p> <p>2h + k - 1 = 0</p> <p>or, k = 1 - 2h.......................(1)</p> <p>Eq<sup>n</sup> of circle is:</p> <p>(x - h)<sup>2</sup> + (y - k)<sup>2</sup> = r<sup>2</sup></p> <p>(x - h)<sup>2</sup> + (y - k)<sup>2</sup> = 5<sup>2</sup>..................(2)</p> <p>The point (4, 3) passes through the eq<sup>n</sup> (2)</p> <p>(4 - h)<sup>2</sup> + (3 - k)<sup>2</sup> = 25</p> <p>or, 16 - 8h + h<sup>2</sup> + 9 - 6k + k<sup>2</sup> = 25</p> <p>or, h<sup>2</sup> - 8h - 6k + k<sup>2</sup> + 25 - 25 = 0</p> <p>or,h<sup>2</sup> - 8h - 6k + k<sup>2</sup>= 0.............................(3)</p> <p>Putting the value of k in eq<sup>n</sup> (3)</p> <p>h<sup>2</sup> - 8h - 6(1 - 2h) + (1 - 2h)<sup>2</sup> = 0</p> <p>or, h<sup>2</sup> - 8h - 6 + 12h + 1 - 4h + 4h<sup>2</sup> = 0</p> <p>or, 5h<sup>2</sup> = 5</p> <p>or, h<sup>2</sup> = \(\frac 55\)</p> <p>or, h<sup>2</sup> = 1</p> <p>∴ h = ± 1</p> <p>Putting the value of k = 1 in eq<sup>n</sup> (1)</p> <p>k = 1 - 2h = 1 - 2× 1 = -1</p> <p>∴ (h, k) = (1, -1)</p> <p>Putting the value of (h, k) in eq<sup>n</sup> (2)</p> <p>(x - h)<sup>2</sup> + (y - k)<sup>2</sup> = 25</p> <p>or, (x - 1)<sup>2</sup> + (y - 1)<sup>2</sup> = 25</p> <p>or, x<sup>2</sup> - 2.x.1 + 1<sup>2</sup> + y<sup>2</sup> - 2.y.1 + 1<sup>2</sup> = 25</p> <p>or, x<sup>2</sup> + y<sup>2</sup> - 2x - 2y + 1 + 1 = 25</p> <p>or,x<sup>2</sup> + y<sup>2</sup> - 2x - 2y = 25 - 2</p> <p>∴x<sup>2</sup> + y<sup>2</sup> - 2x - 2y = 23</p> <p>Putting the value of h = -1 in eq<sup>n</sup> (1)</p> <p>k = 1 - 2h = 1 - 2 × (-1) = 1 + 2 = 3</p> <p>∴ (h, k) = (-1, 3)</p> <p>Putting the value of (h, k) in eq<sup>n</sup> (2)</p> <p>(x - h)<sup>2</sup> + (y - k)<sup>2</sup> = 25</p> <p>or, (x + 1)<sup>2</sup> + (y - 3)<sup>2</sup> = 25</p> <p>or, x<sup>2</sup> + 2x + 1 + y<sup>2</sup> - 6y + 9 = 25</p> <p>or, x<sup>2</sup>+ y<sup>2</sup> + 2x - 6y + 10 = 25</p> <p>or, x<sup>2</sup>+ y<sup>2</sup> + 2x - 6y = 25 - 10</p> <p>∴x<sup>2</sup>+ y<sup>2</sup> + 2x - 6y =15</p> <p>∴ Required equations are:x<sup>2</sup> + y<sup>2</sup> - 2x - 2y = 23 andx<sup>2</sup>+ y<sup>2</sup> + 2x - 6y =15<sub>Ans</sub></p>
Q27:
Find the co-ordinates of the center and the radius of a circle whose equation is:
2x - 6y - x2 - y2 = 1
Type: Long Difficulty: Easy
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Answer: <p>Given equation is:</p> <p>2x - 6y - x<sup>2</sup> - y<sup>2</sup> = 1</p> <p>or, -x<sup>2</sup> + 2x - y<sup>2</sup> - 6y - 1 = 0</p> <p>or, -(x<sup>2</sup> - 2x + y<sup>2</sup> + 6y + 1) = 0</p> <p>or,x<sup>2</sup> - 2x + 1 + y<sup>2</sup> + 6y = 0</p> <p>or, x<sup>2</sup> - 2.x.1 + 1<sup>2</sup> + y<sup>2</sup> + 2.y.3 + 3<sup>2</sup> - 3<sup>2</sup> = 0</p> <p>∴(x - 1)<sup>2</sup> + (y + 3)<sup>2</sup>= 3<sup>2</sup>............................(1)</p> <p>Equation of circle when the centre of circle is (h, k) and radius (r) is:</p> <p>(x - h)<sup>2</sup> + (y - k)<sup>2</sup> = r<sup>2</sup>.................................(2)</p> <p>Comparing (1) and (2)</p> <p>∴ Center of circle (h, k) = (1, -3)</p> <p>and Radius of circle (r) = 3 units<sub>Ans</sub></p>
Q28:
The equation of a circle is: 2x2 + 2y2 - 2x + 6y = 45. Find the co-ordinates of its centre and its radius.
Type: Long Difficulty: Easy
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Answer: <p>Here,</p> <p>2x<sup>2</sup> + 2y<sup>2</sup> - 2x + 6y = 45</p> <p>or, \(\frac {2x^2}2\) + \(\frac {2y^2}2\) - \(\frac {2x}2\) + \(\frac {6y}2\) = \(\frac {45}2\) [\(\because\) Dividing by 2 on both sides.]</p> <p>or, x<sup>2</sup> + y<sup>2</sup> - x + 3y = \(\frac {45}2\)</p> <p>or, x<sup>2</sup>- x + y<sup>2</sup> + 3y = \(\frac {45}2\)</p> <p>or, x<sup>2</sup> - 2.x.\(\frac 12\) + (\(\frac 12)^2\) - (\(\frac 12)^2\) + y<sup>2</sup> + 2.y.\(\frac 32\) + (\(\frac 32)^2\) - (\(\frac 32)^2\) = \(\frac {45}2\)</p> <p>or, (x - \(\frac 12\))<sup>2</sup> + (y + \(\frac 32\))<sup>2</sup> -(\(\frac 14)\) -(\(\frac 94)\) = \(\frac {45}2\)</p> <p>or, (x - \(\frac 12\))<sup>2</sup> + (y + \(\frac 32\))<sup>2</sup>= \(\frac {45}2\) + \(\frac 14\) + \(\frac 94\)</p> <p>or, (x - \(\frac 12\))<sup>2</sup> + (y + \(\frac 32\))<sup>2</sup>= \(\frac {90 + 1 + 9}4\)</p> <p>or, (x - \(\frac 12\))<sup>2</sup> + (y + \(\frac 32\))<sup>2</sup> = \(\frac {100}4\)</p> <p>or, (x - \(\frac 12\))<sup>2</sup> + (y + \(\frac 32\))<sup>2</sup>= (\(\frac {10}4)^2\)..........................(1)</p> <p>The equation of circle is: (x - h)<sup>2</sup>+ (y - k)<sup>2</sup> = r<sup>2</sup>...........................(2)</p> <p>Comparing equation (1) and (2)</p> <p>h = \(\frac 12\)</p> <p>k = \(\frac {-3}2\)</p> <p>r = \(\frac {10}2\) = 5</p> <p>∴ Center of the circle (h, k) = (\(\frac 12\), \(\frac {-3}2\))</p> <p>and Radius (r) = 5 units<sub>Ans</sub></p>
Q29:
Find the equation of the circle passes through the point (2, -1), (6, 1) and (6, 3).
Type: Long Difficulty: Easy
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Answer: <p></p> <figure class="inline-right" style="width: 150px;"><img src="/uploads/n1.jpg" alt="f" width="150" height="150"><figcaption><br></figcaption></figure><p>Let: C(h, k) be the centre of the circle which passes through P(2, -1), Q(6, 1) and R(6, 3).</p> <p>CP<sup>2</sup> = r<sup>2</sup> = (h - 2)<sup>2</sup> + (k + 1)<sup>2</sup></p> <p>CQ<sup>2</sup> = r<sup>2</sup> = (h - 6)<sup>2</sup> + (k - 1)<sup>2</sup></p> <p>CR<sup>2</sup> = r<sup>2</sup> = (h - 6)<sup>2</sup> + (k - 3)<sup>2</sup></p> <p>Now,</p> <p>CP<sup>2</sup>= CQ<sup>2</sup></p> <p>or,(h - 2)<sup>2</sup> + (k + 1)<sup>2 =</sup>(h - 6)<sup>2</sup> + (k - 1)<sup>2</sup></p> <p>or, h<sup>2</sup>- 4h + 4 + k<sup>2</sup> + 2k + 1 = h<sup>2</sup> - 12h + 36 + k<sup>2</sup> - 2k + 1</p> <p>or, 8h + 4k - 32 = 0</p> <p>or, 4(2h + k - 8) = 0</p> <p>or, 2h + k - 8 = 0...........................(1)</p> <p>Also,</p> <p>CP<sup>2</sup> = CR<sup>2</sup></p> <p>or, (h - 2)<sup>2</sup> + (k + 1)<sup>2</sup> = (h - 6)<sup>2</sup> + (k - 3)<sup>2</sup></p> <p>or, h<sup>2</sup> - 4h + 4 + k<sup>2</sup> + 2k + 1 = h<sup>2</sup> - 12h + 36 + k<sup>2</sup> - 6k + 9</p> <p>or, 8h + 8k - 40 = 0</p> <p>or, 8(h + k - 5) = 0</p> <p>or, h + k - 5 = 0..........................(2)</p> <p>Subtracting eq<sup>n</sup> (1) from (2): we get;</p> <table width="167"><tbody><tr><td>h</td> <td>+</td> <td>k</td> <td>-</td> <td>5</td> <td>=</td> <td>0</td> </tr><tr><td>2h</td> <td>+</td> <td>k</td> <td>-</td> <td>8</td> <td>=</td> <td>0</td> </tr><tr><td>-</td> <td>-</td> <td></td> <td>+</td> <td></td> <td></td> <td></td> </tr><tr><td>-h</td> <td></td> <td></td> <td>+</td> <td>3</td> <td>=</td> <td>0</td> </tr></tbody></table><p>or, -h = -3</p> <p>∴ h = 3</p> <p>Putting the value of h in eq<sup>n</sup> (2)</p> <p>h + k - 5 = 0</p> <p>or, 3 + k - 5 = 0</p> <p>or, k - 2 = 0</p> <p>∴ k = 2</p> <p>∴ Centre = (h, k) = (3, 2)</p> <p>and Radius (r) = \(\sqrt {(2 - 3)^2 + (-1 - 2)^2}\) = \(\sqrt {(-1)^2 + (-3)^2}\) = \(\sqrt {10}\)</p> <p>Putting the value of (h, k) and r in eq<sup>n</sup>:</p> <p>(x - h)<sup>2</sup> + (y - k)<sup>2</sup> = r<sup>2</sup></p> <p>or, (x - 3)<sup>2</sup> + (y - 2)<sup>2</sup> = (\(\sqrt {10}\))<sup>2</sup></p> <p>or, x<sup>2</sup> - 6x + 9 + y<sup>2</sup> - 4y + 4 = 10</p> <p>or, x<sup>2</sup> + y<sup>2</sup> - 6x - 4y + 13 = 10</p> <p>or, x<sup>2</sup> + y<sup>2</sup> - 6x - 4y + 13 -10 = 0</p> <p>∴ x<sup>2</sup> + y<sup>2</sup> - 6x - 4y +3 = 0<sub>Ans</sub></p>
Q30:
The line x + y = 2 cuts a circle x2 + y2 = 4 at two points. Find the co-ordinates of the points.
Type: Long Difficulty: Easy
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Answer: <p>Given eq<sup>n</sup> is:</p> <p>x + y = 2</p> <p>or, y = 2 - x......................(1)</p> <p>x<sup>2</sup> + y<sup>2</sup> = 4........................(2)</p> <p>Putting the value of y in equation (2)</p> <p>x<sup>2</sup> + (2 - x)<sup>2</sup> = 4</p> <p>or, x<sup>2</sup> + 4 - 4x + x<sup>2</sup> = 4</p> <p>or, 2x<sup>2</sup> - 4x + 4 - 4 = 0</p> <p>or, 2x<sup>2</sup> - 4x = 0</p> <p>or, 2x (x - 2) = 0</p> <p>Either: 2x = 0 i.e. x = 0</p> <p>Or: x - 2 = 0 i.e. x = 2</p> <p>Putting the value of x in eq<sup>n</sup> (1)</p> <p>If: x = 0</p> <p>y = 2 - 0 =2</p> <p>If: x = 2</p> <p>y = 2 - 2 = 0</p> <p>∴ The required points are: (0, 2) and (2, 0)<sub>Ans</sub></p>
Q31:
The equation of two diameters of a circle passing through the point (3, 4) and x + y = 14 and 2x - y = 4. Find the equation of the circle.
Type: Long Difficulty: Easy
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Answer: <p>Given equation of diameters of the circle are:</p> <p>x + y = 14..........................(1)</p> <p>2x - y = 4...........................(2)</p> <p>Adding equation (1) and (2)</p> <table width="121"><tbody><tr><td>x</td> <td>+</td> <td>y</td> <td>=</td> <td>14</td> </tr><tr><td>2x</td> <td>-</td> <td>y</td> <td>=</td> <td>4</td> </tr><tr><td>3x</td> <td></td> <td></td> <td>=</td> <td>18</td> </tr></tbody></table><p>or, x = \(\frac {18}3\)</p> <p>∴ x = 6</p> <p>Putting the value of x in equation (1)</p> <p>x + y = 14</p> <p>or, 6 + y = 14</p> <p>or, y = 14 - 6</p> <p>∴ y = 8</p> <p>Intersection point of the two diameter equation of the circle is a center of a circle (h, k) = (6, 8)</p> <p>The given circle passes through the point (3, 4)</p> <p>\begin{align*} Radius\; of\; circle &= \sqrt {(x_2 - x_1)^2 + (y_2 - y_1)^2}\\ &=\sqrt {(6 - 3)^2 + (8 - 4)^2}\\ &= \sqrt {3^2 + 4^2}\\ &= \sqrt {9 + 16}\\ &= \sqrt {25}\\ &= 5\; units\\ \end{align*}</p> <p>Equation of the circle whose center is (h, k). Then:</p> <p>(x - h)<sup>2</sup> + (y - k)<sup>2</sup> = r<sup>2</sup></p> <p>or, (x - 6)<sup>2</sup> + (y - 8)<sup>2</sup> = 5<sup>2</sup></p> <p>or, x<sup>2</sup> - 12x + 36 + y<sup>2</sup>- 16y + 64 = 25</p> <p>or, x<sup>2</sup> + y<sup>2</sup> - 12x - 16y + 100 - 25 = 0</p> <p>∴ x<sup>2</sup> + y<sup>2</sup> - 12x - 16y + 75 = 0<sub>Ans</sub></p>
Q32:
A circle has radius 5 units and the equations of its two diameters are 2x - y = 5 and x - 3y + 5 = 0. Find the equation of the circle and show that it passes through the origin.
Type: Long Difficulty: Easy
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Answer: <p>Given equations of two diameters are:</p> <p>2x - y = 5</p> <p>or, y = 2x - 5................(1)</p> <p>x - 3y + 5 = 0...............(2)</p> <p>Putting the value of y in eq<sup>n</sup> (2)</p> <p>x - 3 (2x - 5) + 5 = 0</p> <p>or, x - 6x + 15 + 5 = 0</p> <p>or, -5x = - 20</p> <p>or, x = \(\frac {-20}{-5}\)</p> <p>∴ x = 4</p> <p>Putting the value of x in eq<sup>n</sup> (1)</p> <p>y = 2x - 5</p> <p>or, y = 2× 4 - 5</p> <p>or, y = 8 - 5</p> <p>∴ y = 3</p> <p>The center of the circle (h, k) = (4, 3)</p> <p>Radius of the circle (r) = 5 units</p> <p>Equation of the circle is:</p> <p>(x - h)<sup>2</sup> + (y - k)<sup>2</sup> = r<sup>2</sup></p> <p>or, (x - 4)<sup>2</sup> + (y - 3)<sup>2</sup> = 5<sup>2</sup></p> <p>or, x<sup>2</sup> - 8x + 16 + y<sup>2</sup> - 6y + 9 = 25</p> <p>or, x<sup>2</sup> + y<sup>2</sup> - 8x - 6y + 25 = 25</p> <p>or, x<sup>2</sup> + y<sup>2</sup> - 8x - 6y + 25 -25 = 0</p> <p>∴ x<sup>2</sup> + y<sup>2</sup> - 8x - 6y= 0.............................(3)</p> <p>The point (0, 0) passes through the eq<sup>n</sup> (3)</p> <p>0<sup>2</sup> + 0<sup>2</sup> - 4× 0 - 6× 0 = 0</p> <p>0 = 0</p> <p>∴ The equation x<sup>2</sup> + y<sup>2</sup> - 4x - 6y = 0 passes through origin.<sub>Proved</sub></p> <p></p>
Q33:
If the line joining the points (1, 2) and (3, 6) is the diameter of a circle, find the equation of the circle.
Type: Long Difficulty: Easy
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Answer: <p>Given points are: (1, 2) and (3, 6)</p> <p>Equation of the diameter of a circle is:</p> <p>(x - x<sub>1</sub>) (x - x<sub>2</sub>) + (y - y<sub>1</sub>) (y - y<sub>2</sub>) = 0</p> <p>or, (x - 1) (x - 3) + (y - 2) (y - 6) = 0</p> <p>or, x<sup>2</sup>- x - 3x + 3 + y<sup>2</sup> - 2y - 6y + 12 = 0</p> <p>∴ x<sup>2</sup> + y<sup>2</sup> - 4x - 8y + 15 = 0</p> <p>Hence, the required equation is:x<sup>2</sup> + y<sup>2</sup> - 4x - 8y + 15 = 0<sub>Ans</sub></p>
Q34:
Find the radius of the circle which passes through the points A(-4, -2), B(2, 6) and C(2, -2).
Type: Long Difficulty: Easy
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Answer: <p>Equation of the circle with center at (h, k) and radius (r) is:</p> <p>(x - h)<sup>2</sup> + (y - k)<sup>2</sup> = r<sup>2</sup>.........................(1)</p> <p>Points A(-4, -2), B(2, 6) and C(2, -2) passes through the equation (1)</p> <p>(-4 - h)<sup>2</sup> + (-2 - k)<sup>2</sup> = r<sup>2</sup>.......................(2)</p> <p>(2 - h)<sup>2</sup> + (6 - k)<sup>2</sup> = r<sup>2</sup>..........................(3)</p> <p>(2 - h)<sup>2</sup> + (-2 - k)<sup>2</sup> = r<sup>2</sup>........................(4)</p> <p>Taking equation (2) and (3) and solving:</p> <p>(-4 - h)<sup>2</sup> + (-2 - k)<sup>2</sup> =(2 - h)<sup>2</sup> + (6 - k)<sup>2</sup></p> <p>or, 16 + 8h + h<sup>2</sup> + 4 + 4k + k<sup>2</sup> = 4 - 4h + h<sup>2</sup> + 36 - 12k + k<sup>2</sup></p> <p>or, 8h + 4k + 20 = 40 - 4h - 12k</p> <p>or, 8h + 4h + 4k + 12k = 40 - 20</p> <p>or, 12h + 16k = 20</p> <p>or, 4(3h + 4k) = 20</p> <p>or, 3h + 4k = \(\frac {20}4\)</p> <p>or, 3h + 4k = 5........................(5)</p> <p>Taking equation (3) and (4) and solving:</p> <p>(2 - h)<sup>2</sup> + (6 - k)<sup>2</sup> =(2 - h)<sup>2</sup> + (-2 - k)<sup>2</sup></p> <p>or, 4 - 4h + h<sup>2</sup> + 36 - 12k + k<sup>2</sup> = 4 - 4h + h<sup>2</sup> + 4 + 4k + k<sup>2</sup></p> <p>or, 40 - 4h - 12k = 8 - 4h + 4k</p> <p>or, -12k - 4k = 8 - 40</p> <p>or, -16k = - 32</p> <p>or, k = \(\frac {-32}{-16}\)</p> <p>∴ k = 2</p> <p>Putting the value of k in equation (5)</p> <p>3h + 4× 2 = 5</p> <p>or, 3h + 8 = 5</p> <p>or, 3h = 5 - 8</p> <p>or, h = \(\frac {-3}3\)</p> <p>∴ h = -1</p> <p>Now,</p> <p>Center (h, k) = (-1, 2)</p> <p>One point = B(2, 6)</p> <p>\begin{align*} Radius\; (r) &= \sqrt {(x_2 - x_1)^2 + (y_2 - y_1)^2}\\ &= \sqrt {(2 + 1)^2 + (6 - 2)^2}\\ &= \sqrt {3^2 + 4^2}\\ &= \sqrt {9 + 16}\\ &= \sqrt {25}\\ &= 5\; units\\ \end{align*}</p> <p>∴ Radius of circle is 5 units.<sub>Ans</sub></p>
Q35:
Find the equation of the circle whose center is (2, 1) and which just touches the line 3x + 4y + 1 = 0.
Type: Long Difficulty: Easy
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Answer: <p>Let: r be the radius of the circle.</p> <p>Then: its equation is:</p> <p>(x - 2)<sup>2</sup> + (y - 1)<sup>2</sup> = r<sup>2</sup>..........................(1)</p> <p>The given line will be tangent to the circle. If it's perpendicular distance from the center is equal to the radius.</p> <p>\begin{align*} r &= \begin {vmatrix} \frac {Ax\;+\; By\;+\; C}{\sqrt {A^2 + B^2}} \end {vmatrix}\\ &= \begin {vmatrix} \frac {3 × 2 + (-4) × 1 + 1}{\sqrt {(3)^2 + (-4)^2}}\\ \end {vmatrix}\\ &= \begin {vmatrix} \frac {6-4+1}{\sqrt {9 + 16}}\\ \end {vmatrix}\\ &= \begin {vmatrix} \frac {3}{\sqrt {5}}\\ \end {vmatrix}\\ &= \frac 35\\ \end{align*}</p> <p>Now,</p> <p>Putting the value of r in equation (1)</p> <p>(x - 2)<sup>2</sup> + (y - 1)<sup>2</sup> = (\(\frac 35)^2\)</p> <p>or, x<sup>2</sup> - 4x + 4 + y<sup>2</sup> - 2y + 1 = \(\frac 9{25}\)</p> <p>or, 25x<sup>2</sup> - 100x + 100 + 25y<sup>2</sup> - 50y + 25 - 9 = 0</p> <p>or, 25x<sup>2</sup> + 25y<sup>2</sup> - 100x - 50y + 116 = 0</p> <p>Hence, the required equation is:25x<sup>2</sup> + 25y<sup>2</sup> - 100x - 50y + 116 = 0<sub>Ans</sub></p>
Q36:
Prove that the line 3x + y + 7\(\sqrt {10}\) = 0 is tangent to the circle x2 + y2 - 2x + 6y - 39 = 0.
Type: Long Difficulty: Easy
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Answer: <p>Here,</p> <p>The eq<sup>n</sup> of circle is:</p> <p>x<sup>2</sup> + y<sup>2</sup> - 2x + 6y - 39 = 0</p> <p>or, x<sup>2</sup> - 2.x.1 + 1<sup>2</sup> - 1<sup>2</sup> + y<sup>2</sup> + 2.y.3 + 3<sup>2</sup> - 3<sup>2</sup> - 39 = 0</p> <p>or, (x - 1)<sup>2</sup> + (y + 3)<sup>2</sup> - 1 - 9 - 39 = 0</p> <p>or, (x - 1)<sup>2</sup> + (y + 3)<sup>2</sup> - 49 = 0</p> <p>or, (x - 1)<sup>2</sup> + (y + 3)<sup>2</sup>= 49</p> <p>or, (x - 1)<sup>2</sup> + (y + 3)<sup>2</sup>=7<sup>2</sup>........................(1)</p> <p>The eq<sup>n</sup> of circle is:</p> <p>(x - h)<sup>2</sup> + (y - k)<sup>2</sup> = r<sup>2</sup>........................(2)</p> <p>Comparing equation (1) and (2)</p> <p>Center of circle (h, k) = (1, -3)</p> <p>Radius (r) = 7</p> <p>The given equation of a line is:3x + y + 7\(\sqrt {10}\) = 0</p> <p>The length of the perpendicular drawn from the center (1, -3) on the line3x + y + 7\(\sqrt {10}\) = 0 is:</p> <p>\begin{align*} d &= \frac {3 × 1 - 3 + 7\sqrt {10}}{\sqrt {3^2 + 1^2}}\\ &= \frac {7\sqrt {10}}{\sqrt {10}}\\ &= 7\\ \end{align*}</p> <p>Length of the perpendicular = radius of the circle = 7 units</p> <p>∴ 3x + y + 7\(\sqrt {10}\) = 0 is tangent to the circle.<sub>Proved</sub></p>
Q37:
Find the equation of the tangent to the circle x2 + y2 + 6x - 2y - 15 = 0 at a point (2, 2).
Type: Long Difficulty: Easy
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Answer: <p>Here,</p> <p>The equation of circle is:</p> <p>x<sup>2</sup> + y<sup>2</sup> + 6x - 2y - 15 = 0</p> <p>or, x<sup>2</sup> + 2.x.3 + 3<sup>2</sup> - 3<sup>2</sup> + y<sup>2</sup> - 2.y.1 + 1<sup>2</sup> - 1<sup>2</sup> - 15 = 0</p> <p>or, (x + 2)<sup>2</sup> + (y - 1)<sup>2</sup> - 9 - 1 - 15 = 0</p> <p>or, (x + 2)<sup>2</sup> + (y - 1)<sup>2</sup> - 25 = 0</p> <p>or, (x + 2)<sup>2</sup> + (y - 1)<sup>2</sup>= 25</p> <p>or, (x + 2)<sup>2</sup> + (y - 1)<sup>2</sup>=5<sup>2</sup>........................(1)</p> <p>The eq<sup>n</sup> of circle is:</p> <p>(x - h)<sup>2</sup> + (y - k)<sup>2</sup> = r<sup>2</sup>......................(2)</p> <p>Comparing (1) and (2)</p> <p>Center (h, k) = (-3, 1) and radius (r) = 5</p> <p>The point of contact is: A(2, 2)</p> <p>Slope of OA = \(\frac {y_2 - y_1}{x_2 - x_1}\) = \(\frac {2 -1}{2 + 3}\) = \(\frac 15\)</p> <p>Slope of the tangent PQ (m) = -5</p> <p>Equation of the tangent PQ at (2, 2) is:</p> <p>y - y<sub>1</sub> = m(x - x<sub>1</sub>)</p> <p>or, y - 2 = -5(x - 1)</p> <p>or, y - 2 = -5x + 5</p> <p>or, 5x + y = 5 + 2</p> <p>∴ 5x + y = 7<sub>Ans</sub></p>
Q38:
Find the equation of the circle which passes through the points (3, 2) and (5, 4) and the center lies on the line 3x - 2y = 1.
Type: Long Difficulty: Easy
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Answer: <p>Let: The center of circle is be (h, k) and radius be r.</p> <p>Equation of circle is:</p> <p>(x - h)<sup>2</sup> + (y - h)<sup>2</sup> = r<sup>2</sup>....................(1)</p> <p>The point (3, 2) and (5, 4) passes through eq<sup>n</sup> (1)</p> <p>(3 - h)<sup>2</sup> + (2 - k)<sup>2</sup> = r<sup>2</sup>....................(2)</p> <p>(3 - h)<sup>2</sup> + (4 - k)<sup>2</sup> = r<sup>2</sup>....................(3)</p> <p>From (2) and (3)</p> <p>(3 - h)<sup>2</sup> + (2 - k)<sup>2</sup> =(3 - h)<sup>2</sup> + (4 - k)<sup>2</sup></p> <p>or, 9 - 6h + h<sup>2</sup> + 4 - 4k + k<sup>2</sup> = 25 - 10h + h<sup>2</sup> + 16 - 8k + k<sup>2</sup></p> <p>or, 4h + 4k - 28 = 0</p> <p>or, 4(h + k - 7) = 0</p> <p>or, h + k - 7 = 0</p> <p>or, h = 7 - k.................(4)</p> <p>The point (h, k) lines in the line:</p> <p>3x - 2y = 1</p> <p>3h - 2k = 1...................(5)</p> <p>Putting the value of (h, k) in eq<sup>n</sup>(5)</p> <p>3(7 - k) - 2k = 1</p> <p>or, 21 - 3k - 2k = 1</p> <p>or, 21 - 5k = 1</p> <p>or, -5k = 1 - 21</p> <p>or, -5k = -20</p> <p>or, k = \(\frac {-20}{-5}\)</p> <p>∴ k = 4</p> <p>Putting the value of k in eq<sup>n</sup> (4)</p> <p>h = 7 - k = 7 - 4 = 3</p> <p>Putting the value of (h, k) in eq<sup>n</sup> (2)</p> <p>(3 - 3)<sup>2</sup> + (2 - 4)<sup>2</sup> = r<sup>2</sup></p> <p>or, 0 + (-2)<sup>2</sup> = r<sup>2</sup></p> <p>or, r<sup>2</sup> = 4</p> <p>∴ r = 2</p> <p>Putting the value of (h, k) in eq<sup>n</sup> (1)</p> <p>(x - 3)<sup>2</sup> + (y - 4)<sup>2</sup> = 2<sup>2</sup></p> <p>or, x<sup>2</sup> - 6x + 9 + y<sup>2</sup> - 8y + 16 = 4</p> <p>or, x<sup>2</sup> + y<sup>2</sup> - 6x - 8y + 25 - 4 = 0</p> <p>∴ x<sup>2</sup> + y<sup>2</sup> - 6x - 8y + 21 = 0<sub> Ans</sub></p> <p></p>
Q39:
Find the equation of the circle whose center is the point of intersection of x + 2y - 1 = 0 and 2x - y - 7 = 0 and which passes through (3, 1).
Type: Long Difficulty: Easy
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Answer: <p>Here,</p> <p>x + 2y - 1 = 0.....................(1)</p> <p>2x - y - 7 = 0......................(2)</p> <p>Eq<sup>n</sup> (1) is multiplied by 2 and adding with eq<sup>n</sup> (1)</p> <table width="192"><tbody><tr><td>4x</td> <td>-</td> <td>2y</td> <td>-</td> <td>14</td> <td>=</td> <td>0</td> </tr><tr><td>x</td> <td>+</td> <td>2y</td> <td>-</td> <td>1</td> <td>=</td> <td>0</td> </tr><tr><td>5x</td> <td></td> <td></td> <td>-</td> <td>15</td> <td>=</td> <td>0</td> </tr></tbody></table><p>or, 5x = 15</p> <p>or, x = \(\frac {15}5\)</p> <p>∴ x = 3</p> <p>Putting the value of x in eq<sup>n</sup> (1)</p> <p>x + 2y - 1 = 0</p> <p>or, 3 + 2y - 1 = 0</p> <p>or, 2y = -2</p> <p>or, y = -\(\frac 22\)</p> <p>∴ y = -1</p> <p>The center of circle = (h, k) = (3, -1)</p> <p>\begin{align*} The\; radius\; of\; the\; circle\; (r) &= \sqrt {(3 - 3)^2 + (1 - (-1))^2}\\ &= \sqrt {0 + (-2)^2}\\ &= \sqrt 4\\ &= 2\; units \end{align*}</p> <p>The eq<sup>n</sup> of circle is: (x - h)<sup>2</sup> + (y - k)<sup>2</sup> = r<sup>2</sup>.........................(3)</p> <p>Putting the value of (h, k) and r in eq<sup>n</sup> (2)</p> <p>(x - 3)<sup>2</sup> + (y + 1)<sup>2</sup> = 2<sup>2</sup></p> <p>or, x<sup>2</sup>- 6x + 9 + y<sup>2</sup>+ 2y + 1 = 4</p> <p>or, x<sup>2</sup> + y<sup>2</sup> - 6x + 2y + 10 - 4 = 0</p> <p>∴ x<sup>2</sup> + y<sup>2</sup> - 6x + 2y + 6 = 0<sub>Ans</sub></p>
Q40:
The center of a circle which passes through the origin and the point (4, 2) lies on the line x + y = 1. Find the equation of the circle.
Type: Long Difficulty: Easy
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Answer: <p>Let:</p> <p>Center of the circle be (h, k) and radius be r.</p> <p>Equation of the circle is:</p> <p>(x - h)<sup>2</sup> + (y - k)<sup>2</sup> = r<sup>2</sup>......................(1)</p> <p>The point (0, 0) and (4, 2) are passing through the eq<sup>n</sup> (1)</p> <p>(0 - h)<sup>2</sup> + (0 - k)<sup>2</sup> = r<sup>2</sup></p> <p>h<sup>2</sup> + k<sup>2</sup> = r<sup>2</sup>.................(2)</p> <p>(4 - h)<sup>2</sup> + (2 - k)<sup>2</sup> = r<sup>2</sup>.......................(3)</p> <p>From eq<sup>n</sup> (2) and (3)</p> <p>h<sup>2</sup> + k<sup>2</sup> =(4 - h)<sup>2</sup> + (2 - k)<sup>2</sup></p> <p>or, h<sup>2</sup> + k<sup>2</sup> = 16 - 8h + h<sup>2</sup> + 4 - 4k + k<sup>2</sup></p> <p>or, 8h + 4k - 20 = 0</p> <p>or, 4(2h + k - 5) = 0</p> <p>or, 2h + k - 5 = 0</p> <p>or, k = 5 - 2h...........................(4)</p> <p>The point (h, k) lies on the line:</p> <p>x + y = 1</p> <p>h + k = 1........................(5)</p> <p>Putting the value of k in eq<sup>n</sup> (5)</p> <p>h + 5 - 2h = 1</p> <p>or, 5 - 1 = h</p> <p>∴h = 4</p> <p>Putting the value of h in eq<sup>n</sup> (4)</p> <p>k = 5 - 2h = 5 - 2× 4 = 5 - 8 = -3</p> <p>Putting the value of (h, k) = (4, -3) in eq<sup>n</sup> (2)</p> <p>h<sup>2</sup> + k<sup>2</sup> = r<sup>2</sup></p> <p>or, r<sup>2</sup> = 4<sup>2</sup> + (-3)<sup>2</sup></p> <p>or, r<sup>2</sup> = 16 + 9</p> <p>or, r<sup>2</sup> = 25</p> <p>∴ r = 5</p> <p>Putting the value of (h, k) and r in the eq<sup>n</sup></p> <p>(x - h)<sup>2</sup> + (y - k)<sup>2</sup> = r<sup>2</sup></p> <p>or, (x - 4)<sup>2</sup> + (y + 3)<sup>2</sup> = 5<sup>2</sup></p> <p>or, x<sup>2</sup>- 8x + 16 + y<sup>2</sup> + 6y + 9 = 25</p> <p>or, x<sup>2</sup> + y<sup>2</sup> - 8x + 6y + 25 - 25 = 0</p> <p>∴x<sup>2</sup> + y<sup>2</sup> - 8x + 6y = 0<sub>Ans</sub></p>
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Meiosis Cell Division
It is reductional division, which takes place in the reproductive cell. It is a longer process in comparison to mitosis. It completes in the following steps:
Meiosis I:
It is further divided into:
- Prophase I: It is the longest stage of cell division, which is further divided into five sub-stages:
- Leptotene: The nucleus increases in size and prepares the cell for division. Chromosomes are coiled with each other and they are thelong thread like in structure. Such chromosomes are called () like chromosome.
- Zygotene: In this stage, the chromatid threads having similar structure, function etc. but donated by opposite sex (i.e. mother and father) starts to make their pair, such pair of the chromosome are called homologous chromosome. In this stage, the centromere starts to appear which divides the chromosome into two arms, such chromosome is called condensed chromosome. The process of formation of thehomologous chromosome is called synapse.
- Pachytene: The chromosome splits lengthwise except centromere and the resulting chromosomes are called tetrad chromosomes. In late pachytene, the crossing over process will initiate.
- Diplotene: In this stage, the exchange of genetic materials between paired chromosomes occurs. This process is called crossing over. Crossing over is important for evolution as it helps to bring variation. The point where the paired chromosomes attach to each other during crossing over it is called chaismata.
- Diakinesis: Centriole divides into two and starts to move towards opposite part of the cell. Chromosomes are much thick and condensed. In late diakinesis except centriole chromosomes and cell membrane, other parts start to disappear.
- Metaphase I: Centrioles reach to the opposite poles of the cell. Chromosomes are arranged in theequatorial plane of the cell in two rows. Spindlefibre appears.
- Anaphase I:: Each paired chromosome starts to move towards the opposite poles of the cell by the help of contraction of spindlefibre and the repulsion between chromosomes. Chromosomes appear in the shape of I, J,U and V.
- Telophase I:: The paired chromosomes reach to the opposite poles of the cell. They are also surrounded by a nuclear membrane. Hence diploid daughter nuclei are formed. The cellularconstriction becomes more andmore deep and finally divides the cell into two equal halves.
Meiosis II
The two daughter cells that are produced at the end of telophase I undergo meiosis II. The process of meiosis II includes:
- Prophase II: The centriole divides in two and starts to move towards the opposite pole of the cell. Chromosomes are shorter, thicker and distinct.
- Metaphase II: Centrioles have reached their opposite poles of the cell. Chromosomes are arranged in an equatorial plane of the cell. Spindle fibre appears.
- Anaphase II: Each chromosome splits lengthwise through centromere and produces sister chromatids. Each chromatid, start to move towards the opposite pole of the cell by the contraction of spindle fibre and repulsion between them.
- Telophase II: Chromatids reach to opposite poles of the cell. They are surrounded by the nuclear membrane. Hence, four haploid daughter cells are produced at the end of meiosis II. The cellularconstriction divides the cell into four equal parts. All cellular parts reappear.
Importance Of meiosis cell division:
- Variation: Crossing over that starts in pachytene and ends at diplotene of prophase I help to exchange the genetic material between chromosomes that creates variations among offspring.
- Sexual reproduction: Meiosis cell division occurs only in reproductive cells and produces gametes. So, it helps to carry out sexual reproduction.
- Genetic stability: The offsprings, which are produced after sexual reproduction, are similar to their parents in many respects. Therefore, meiosis helps to keep genetic stability.
Lesson
Cell Division
Subject
Science
Grade
Grade 10
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