JEE Main 2026 Apr 5 Shift 2, Mathematics Q11: Involving Two or More Circles

JEE Main2026Apr 5, Shift 2Mathematics

Let the point $P$ be the vertex of the parabola $y = x^{2} - 6 x + 12$ . If a line passing through the point $P$ intersects the circle $x^{2} + y^{2} - 2 x - 4 y + 3 = 0$ at the points $R$ and $S$ , then the maximum value of $(P R + P S)^{2}$ is:

A
10
B
20
C
25
D
5

Solution

Complete the square: $y = (x - 3)^{2} + 3$ , so the vertex is $P (3, 3)$ .

The circle has centre $C (1, 2)$ and radius $r = 1 + 4 - 3 = 2$ .

$P C = (3 - 1)^{2} + (3 - 2)^{2} = 5$ .

For a chord through $P$ , $(P R + P S)$ is maximised when the chord is the diameter through $P$ (the chord of longest length visible from $P$ along the line through $C$ ). Then $P R + P S = (P C - r) + (P C + r) = 2 P C$ .

$(P R + P S)_{m a x}^{2} = (2 P C)^{2} = 4 \cdot 5 = 20$ .

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JEE Main 2026 Apr 5 Shift 2, Mathematics Q11: Involving Two or More Circles

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