Let C be a circle passing through the points A(2, -1) and B(3, 4). The line segment AB is not a diameter of C. If r is the radius of C and its centre lies on the circle <math> <mrow> <msup> <mrow> <mrow> <mo>(</mo> <mrow> <mi>x</mi> <mo>−</mo> <mn>5</mn> </mrow> <mo>)</mo> </mrow> </mrow> <mrow> <mn>5</mn> </mrow> </msup> <mo>+</mo> <msup> <mrow> <mrow> <mo>(</mo> <mrow> <mi>y</mi> <mo>−</mo> <mn>1</mn> </mrow> <mo>)</mo> </mrow> </mrow> <mrow> <mn>2</mn> </mrow> </msup> <mo>=</mo> <mfrac> <mrow> <mn>1</mn> <mn>3</mn> </mrow> <mrow> <mn>2</mn> </mrow> </mfrac> </mrow> </math> , then r2 is equal to:

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Conic Sections Ncert Solutions Maths class 11th Maths Class 11th

Let C be a circle passing through the points A(2, -1) and B(3, 4). The line segment AB is not a diameter of C. If r is the radius of C and its centre lies on the circle ${(x - 5)}^{5} + {(y - 1)}^{2} = \frac{13}{2}$ , then r² is equal to:

Option 1 -

Option 2 -

$\frac{65}{2}$

Option 3 -

$\frac{61}{2}$

Option 4 -

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1 Answer

Answered by

Raj Pandey | Contributor-Level 9

a month ago

Correct Option - 2

Detailed Solution:

Equation of perpendicular bisector of AB is

$y - \frac{3}{2} = - \frac{1}{5} (x - \frac{5}{2}) \Rightarrow x + 5 y = 10$

Solving it with equation of given circle

${(x - 5)}^{2} + {(\frac{10 - x}{5} - 1)}^{2} = \frac{13}{2}$

$\Rightarrow x - 5 = \pm \frac{5}{2} \Rightarrow x = \frac{5}{2} o r \frac{15}{2}$

But $x \neq \frac{5}{2}$

because AB is not the diameter.

So, centre will be

$(\frac{15}{2}, \frac{1}{2})$

Now,

$r^{2} = {(\frac{15}{2} - 2)}^{2} + {(\frac{1}{2} + 1)}^{2} = \frac{65}{2}$

Equation of perpendicular bisector of AB is <math> <mrow> <mi>y</mi> <mo>−</mo> <mfrac> <mrow> <mn>3</mn> </mrow> <mrow> <mn>2</mn> </mrow> </mfrac> <mo>=</mo> <mo>−</mo> <mfrac> <mrow> <mn>1</mn> </mrow> <mrow> <mn>5</mn> </mrow> </mfrac> <mrow> <mo> (</mo> <mrow> <mi>x</mi> <mo>−</mo> <mfrac> <mrow> <mn>5</mn> </mrow> <mrow> <mn>2</mn> </mrow> </mfrac> </mrow> <mo>)</mo> </mrow> <mo>⇒</mo> <mi>x</mi> <mo>+</mo> <mn>5</mn> <mi>y</mi> <mo>=</mo> <mn>1</mn> <mn>0</mn> </mrow> </math> Solving it with equation of given circle <math> <mrow> <msup> <mrow> <mrow> <mo> (</mo> <mrow> <mi>x</mi> <mo>−</mo> <mn>5</mn> </mrow> <mo>)</mo> </mrow> </mrow> <mrow> <mn>2</mn> </mrow> </msup> <mo>+</mo> <msup> <mrow> <mrow> <mo> (</mo> <mrow> <mfrac> <mrow> <mn>1</mn> <mn>0</mn> <mo>−</mo> <mi>x</mi> </mrow> <mrow> <mn>5</mn> </mrow> </mfrac> <mo>−</mo> <mn>1</mn> </mrow> <mo>)</mo> </mrow> </mrow> <mrow> <mn>2</mn> </mrow> </msup> <mo>=</mo> <mfrac> <mrow> <mn>1</mn> <mn>3</mn> </mrow> <mrow> <mn>2</mn> </mrow> </mfrac> </mrow> </math> <math> <mrow> <mo>⇒</mo> <mi>x</mi> <mo>−</mo> <mn>5</mn> <mo>=</mo> <mo>±</mo> <mfrac> <mrow> <mn>5</mn> </mrow> <mrow> <mn>2</mn> </mrow> </mfrac> <mo>⇒</mo> <mi>x</mi> <mo>=</mo> <mfrac> <mrow> <mn>5</mn> </mrow> <mrow> <mn>2</mn> </mrow> </mfrac> <mi>o</mi> <mi>r</mi> <mtext> </mtext> <mtext> </mtext> <mfrac> <mrow> <mn>1</mn> <mn>5</mn> </mrow> <mrow> <mn>2</mn> </mrow> </mfrac> </mrow> </math> But <math> <mrow> <mi>x</mi> <mo>≠</mo> <mfrac> <mrow> <mn>5</mn> </mrow> <mrow> <mn>2</mn> </mrow> </mfrac> </mrow> </math> because AB is not the diameter.So, centre will be <math> <mrow> <mrow> <mo> (</mo> <mrow> <mfrac> <mrow> <mn>1</mn> <mn>5</mn> </mrow> <mrow> <mn>2</mn> </mrow> </mfrac> <mo>, </mo> <mfrac> <mrow> <mn>1</mn> </mrow> <mrow> <mn>2</mn> </mrow> </mfrac> </mrow> <mo>)</mo> </mrow> </mrow> </math> Now, <math> <mrow> <msup> <mrow> <mi>r</mi> </mrow> <mrow> <mn>2</mn> </mrow> </msup> <mo>=</mo> <msup> <mrow> <mrow> <mo> (</mo> <mrow> <mfrac> <mrow> <mn>1</mn> <mn>5</mn> </mrow> <mrow> <mn>2</mn> </mrow> </mfrac> <mo>−</mo> <mn>2</mn> </mrow> <mo>)</mo> </mrow> </mrow> <mrow> <mn>2</mn> </mrow> </msup> <mo>+</mo> <msup> <mrow> <mrow> <mo> (</mo> <mrow> <mfrac> <mrow> <mn>1</mn> </mrow> <mrow> <mn>2</mn> </mrow> </mfrac> <mo>+</mo> <mn>1</mn> </mrow> <mo>)</mo> </mrow> </mrow> <mrow> <mn>2</mn> </mrow> </msup> <mo>=</mo> <mfrac> <mrow> <mn>6</mn> <mn>5</mn> </mrow> <mrow> <mn>2</mn> </mrow> </mfrac> </mrow> </math>

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Let C be a circle passing through the points A(2, -1) and B(3, 4). The line segment AB is not a diameter of C. If r is the radius of C and its centre lies on the circle ${(x - 5)}^{5} + {(y - 1)}^{2} = \frac{13}{2}$ , then r² is equal to:

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Let C be a circle passing through the points A(2, -1) and B(3, 4). The line segment AB is not a diameter of C. If r is the radius of C and its centre lies on the circle ( x − 5 ) 5 + ( y − 1 ) 2 = 1 3 2 , then r2 is equal to:

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Let C be a circle passing through the points A(2, -1) and B(3, 4). The line segment AB is not a diameter of C. If r is the radius of C and its centre lies on the circle ${(x - 5)}^{5} + {(y - 1)}^{2} = \frac{13}{2}$ , then r² is equal to: