Maths Q11 – JEE-MAIN 2026

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:

Clue 1 of 3: Geometric Interpretation

Understand the relationship between the external point P, the circle, and the chord RS formed by the intersecting line.

🧠 Full Solution Path

Step 1: Find the vertex of the parabola and the center/radius of the circle✦ Active

The parabola equation is $y = x^{2} - 6 x + 12$ . Completing the square, we get $y = (x - 3)^{2} + 3$ . Thus, the vertex P is $(3, 3)$ .

The circle equation is $x^{2} + y^{2} - 2 x - 4 y + 3 = 0$ . Completing the square for x and y terms, we get $(x - 1)^{2} + (y - 2)^{2} = 2$ . Thus, the center C is $(1, 2)$ and the radius $r = \sqrt{2}$ .

Step 2: Relate PR+PS to geometric distances○ Expand

Calculate the distance between P and C: $P C = \sqrt{(3 - 1)^{2} + (3 - 2)^{2}} = \sqrt{2^{2} + 1^{2}} = \sqrt{4 + 1} = \sqrt{5}$ . Since $P C = \sqrt{5} > r = \sqrt{2}$ , point P is outside the circle.

Let a line passing through P intersect the circle at R and S. Let M be the midpoint of the chord RS. Since P is external to the circle, P is external to the segment RS. Therefore, $P R + P S = 2 P M$ .

In the right-angled triangle $△ P M C$ (where CM is perpendicular to RS), we have $P C^{2} = P M^{2} + C M^{2}$ . Substituting the value of $P C^{2}$ , we get $5 = P M^{2} + C M^{2}$ .

💡 Teacher's Secret Hint

Remember that for an external point P and a chord RS, $P R + P S = 2 P M$ where M is the midpoint of RS.

Step 3: Maximize

(P R + P S)^{2}

○ Expand

We need to maximize $(P R + P S)^{2} = (2 P M)^{2} = 4 P M^{2}$ .

From the relation $5 = P M^{2} + C M^{2}$ , we can write $P M^{2} = 5 - C M^{2}$ . To maximize $P M^{2}$ , we must minimize $C M^{2}$ .

The minimum value of $C M$ is $0$ , which occurs when the line passing through P also passes through the center C of the circle. In this case, M coincides with C.

When $C M = 0$ , $P M^{2} = 5$ . Therefore, the maximum value of $(P R + P S)^{2} = 4 \times 5 = 20$ .

💡 Teacher's Secret Hint

The distance from the center to a chord is minimized when the chord passes through the center itself.

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Maths Question 11 – JEE-MAIN 2026

Clue 1 of 3: Geometric Interpretation

🧠 Full Solution Path