Mathematics · Circles

JEE Advanced 2023 — Paper 2 — Question 13

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 $1<\mathrm{r}<3$ . Two distinct common tangents $P Q$ and $S T$ of $C_{1}$ and $C_{2}$ are drawn. The tangent $P Q$ touches $C_{1}$ at $P$ and $C_{2}$ at $Q$ . The tangent $S T$ touches $C_{1}$ at $S$ and $C_{2}$ at $T$ . Mid points of the line segments $P Q$ and $S T$ are joined to form a line which meets the $x$ -axis at a point $B$ . If $A B=\sqrt{5}$ , then the value of $r^{2}$ is

Answer: 2

Numerical answer — enter this value.

Step-by-step solution

Line joining $\mathrm{M}_{1} \mathrm{M}_{2}$ will be radical axis of two circles $C_{1}: x^{2}+y^{2}-1=0$ $C_{2}:(x-4)^{2}+(y-1)^{2}=r^{2}$ $x^{2}+y^{2}-8 x-2 y+\left(17-r^{2}\right)=0$ Line $\mathrm{M}_{1} \mathrm{M}_{2}:\left(\mathrm{x}^{2}+\mathrm{y}^{2}-1\right)-\left(\mathrm{x}^{2}+\mathrm{y}^{2}-8 \mathrm{x}-2 \mathrm{y}+\left(17-\mathrm{r}^{2}\right)\right)=0$ $8 x+2 y+\left(r^{2}-18\right)=0$ Point $B=\left(\frac{18-r^{2}}{8}, 0\right)$ $\mathrm{AB}=\sqrt{5}$ $\sqrt{\left(\frac{18-\mathrm{r}^{2}}{8}-4\right)^{2}+(0-1)^{2}}=\sqrt{5}$ $\left(\frac{r^{2}+14}{8}\right)^{2}=4$ $\frac{r^{2}+14}{8}=2 \Rightarrow r^{2}=2$

Answer key and solution verified before publishing.

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Exam: JEE Advanced 2023
Paper: Paper 2
Subject: Mathematics
Chapter: Circles
Topic: Tangent & Normal , pair of tangents to circle , chord of contact