Test Index

Circle

Section: Mathematics

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Question : 3 of 27

Marks: +1, -0

Let

C_1

be the circle of radius 1 with center at the origin. Let

C_2

be the circle of radius

r

with center at the point

A=(4,1)

1< r< 3

. Two distinct common tangents

P Q

S T

C_1

C_2

are drawn. The tangent

P Q

C_1

P

C_2

Q

S T

C_1

S

C_2

T

. Midpoints of the line segments

P Q

S T

are joined to form a line which meets the

x

-axis at a point

B

A B=\sqrt{5}

, then the value of

r^2

[JEE Adv 2023 P2]

Your Answer:

Solution:

Let

M

and

N

be midpoints of

P Q

and

S T

respectively.

\Rightarrow M N

is a radical axis of two circles

C_1: x^2+y^2=1

.....(i)

C_2:(x-4)^2+(y-1)^2=r^2

\Rightarrow x^2+y^2-8 x-2 y+17-r^2=0

....(ii)

From (i) and (ii);

Equation of

M N: 8 x+2 y-18+r^2=0

\Rightarrow B

is on

x

-axis

\Rightarrow B \left(\; \frac{18-r^2}{8} , 0\right)

A B=\sqrt{5}

\sqrt{ \left( \; \frac{18-r^2}{8} - 4 \right)^2 + 1 } = \sqrt{5}

A B=\sqrt{5}

\sqrt{ \left( \; \frac{18-r^2}{8} - 4 \right)^2 + 1 } = \sqrt{5}

(By distance formed a)

\Rightarrow

On solving

r^2=2

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