Let $\mathrm{C}_{1}$ be the circle of radius 1 with center at the origin. Let $\mathrm{C}_{2}$ be the circle of radius r with center at the point $\mathrm{A}=(4,1)$, where $1<\mathrm{r}<3$. Two distinct common tangents PQ and ST of $\mathrm{C}_{1}$ and $\mathrm{C}_{2}$ are drawn. The tangent PQ touches $\mathrm{C}_{1}$ at P and $\mathrm{C}_{2}$ at Q. The tangent ST touches $\mathrm{C}_{1}$ at S and $\mathrm{C}_{2}$ at T. Mid points of the line segments PQ and ST are joined to form a line which meets the x-axis at a point B . If $\mathrm{AB}=\sqrt{5}$, then the value of $\mathrm{r}^{2}$ is

JEE Advanced · 2023 · 04 Jun 2023 · Paper 2 · Circle

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)

, where

1 < r < 3

. Two distinct common tangents PQ and ST of

C_{1}

and

C_{2}

are drawn. The tangent PQ touches

C_{1}

at P and

C_{2}

at Q. The tangent ST touches

C_{1}

at S and

C_{2}

at T. Mid points of the line segments PQ and ST are joined to form a line which meets the x-axis at a point B . If

AB = 5

, then the value of

r^{2}

Answer: 2

The radical axis of two circles is the locus of points having equal power with respect to both circles.

For

C_{1} : x^{2} + y^{2} = 1

and

C_{2} : (x - 4)^{2} + (y - 1)^{2} = r^{2}

:

Radical axis:

x^{2} + y^{2} - 1 = (x - 4)^{2} + (y - 1)^{2} - r^{2}

Simplifying:

8 x + 2 y = 18 - r^{2}

This line meets the x-axis where

y = 0

x = \frac{18 - r ^{2}}{8}

B = (\frac{18 - r ^{2}}{8}, 0)

and

A = (4, 1)

.

Given

A B = 5

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

(\frac{32 - 18 + r ^{2}}{8})^{2} + 1 = 5

(\frac{14 + r ^{2}}{8})^{2} = 4

\frac{14 + r ^{2}}{8} = \pm 2

Taking the positive value:

14 + r^{2} = 16

, so

r^{2} = 2

JEE Advanced · 2023 · 04 Jun 2023 · Paper 2 · Integer answer