JEE Advanced 2023 — Paper 2 — MATH-III solution

JEE Advanced2023Paper 2MATH-III

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 $P Q$ 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 $A B = 5$ , then the value of $r^{2}$ is

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Solution

For any pair of circles, the midpoints of the common tangent segments lie on the radical axis of the two circles. So the line joining these midpoints is the radical axis.

The circles are $C_{1} : x^{2} + y^{2} - 1 = 0$ and $C_{2} : (x - 4)^{2} + (y - 1)^{2} - r^{2} = 0$ , i.e. $x^{2} + y^{2} - 8 x - 2 y + 17 - r^{2} = 0$ .

Subtracting gives the radical axis $8 x + 2 y = 18 - r^{2} .$

It meets the $x$ -axis at $B = (\frac{18 - r ^{2}}{8}, 0)$ . With $A = (4, 1)$ ,

$A B^{2} = (\frac{18 - r ^{2}}{8} - 4)^{2} + 1 = 5 \Rightarrow (\frac{- 14 - r ^{2}}{8})^{2} ?$

Recompute: $\frac{18 - r ^{2}}{8} - 4 = \frac{- 14 - r ^{2}}{8}$ , and squaring gives $\frac{( 14 + r ^{2} ) ^{2}}{64} + 1 = 5$ , hence $(14 + r^{2})^{2} = 256$ , so $14 + r^{2} = 16$ (taking the positive root since $r^{2} > 0$ ). Therefore $r^{2} = 2.$

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