JEE Advanced 2023 Paper 2 · Q13 · Circles

Question

Let $C_1$ be the circle of radius $1$ with centre at the origin. Let $C_2$ be the circle of radius $r$ with centre 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 $AB = \sqrt{5}$, then the value of $r^2$ is ___.

SolveKar AI · Accepted Answer

The line joining mid-points of $PQ$ and $ST$ is the radical axis of $C_1$ and $C_2$. Equation: $S_1 - S_2 = 0$, i.e. $(x^2+y^2-1) - ((x-4)^2 + (y-1)^2 - r^2) = 0$. Simplifying: $8x + 2y - 18 + r^2 = 0$. At $y = 0$: $x = (18 - r^2)/8$. So $B = ((18-r^2)/8, 0)$. $AB^2 = ((18-r^2)/8 - 4)^2 + 1 = ((18 - r^2 - 32)/8)^2 + 1 = ((r^2 + 14)/8)^2 + 1$. Wait — sign: $((18-r^2)/8 - 4) = (18 - r^2 - 32)/8 = -(14 + r^2)/8$. So $AB^2 = (14+r^2)^2/64 + 1 = 5$, giving $(14+r^2)^2 = 256 \Rightarrow 14 + r^2 = 16 \Rightarrow r^2 = 2$.

JEE Advanced 2023 Paper 2 · Q13 · Circles

Solution