JEE Main 2026 · 5 Apr · Shift 2 · Mathematics Q11

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MathematicsSingle correct+4 / -1Question 11 of 75

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

Answer and worked solution

Official answer: B

Points. The vertex is $P(3,3)$ , since $y=(x-3)^2+3$ . The circle has centre $C(1,2)$ and radius $\sqrt2$ , and $PC=\sqrt5$ , so P is outside it.

Chord through P. R and S are on the same side of P. If the line makes angle θ with PC, then $PR+PS=2\,PC\cos\theta\le2\sqrt5$ . Equality holds for the line through the centre.

Maximum: $(2\sqrt5)^2=20$ .

(B)

Topics:Equation of a circle Parabola

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JEE Main 2026 · 5 Apr · Shift 2, Mathematics question 11