JEE Main 2026 — Mathematics Coordinate Geometry

The equation of the parabola is $y = x^2 - 6x + 12$y=x2−6x+12, which can be rewritten as $y - 3 = (x - 3)^2$y−3=(x−3)2.

The vertex of the parabola is $P(3, 3)$P(3,3).

The equation of the circle is $x^2 + y^2 - 2x - 4y + 3 = 0$x2+y2−2x−4y+3=0, which can be rewritten as $(x - 1)^2 + (y - 2)^2 = 2$(x−1)2+(y−2)2=2.

The center of the circle is $C(1, 2)$C(1,2) and its radius is $r = \sqrt{2}$r=2​.

The distance between the point $P$P and the center $C$C is:

$$PC = \sqrt{(3 - 1)^2 + (3 - 2)^2} = \sqrt{4 + 1} = \sqrt{5}$$PC=(3−1)2+(3−2)2​=4+1​=5​

Since $PC = \sqrt{5} > r = \sqrt{2}$PC=5​>r=2​, the point $P$P lies outside the circle.

Let the line through $P$P intersect the circle at $R$R and $S$S, and let $M$M be the midpoint of the chord $RS$RS. The perpendicular distance from the center $C$C to the chord is $CM$CM.

Using the right-angled triangle $\triangle PCM$△PCM, the distance from $P$P to the midpoint $M$M is $PM = \sqrt{PC^2 - CM^2}$PM=PC2−CM2​.

The lengths of the segments are $PR = PM - RM$PR=PM−RM and $PS = PM + RM$PS=PM+RM.

Therefore, the sum of the distances is:

$$PR + PS = (PM - RM) + (PM + RM) = 2PM = 2\sqrt{PC^2 - CM^2}$$PR+PS=(PM−RM)+(PM+RM)=2PM=2PC2−CM2​

To maximize $(PR + PS)^2$(PR+PS)2, we need to maximize $PC^2 - CM^2$PC2−CM2. This occurs when $CM$CM is minimized, which means $CM = 0$CM=0. This happens when the line passes through the center $C$C of the circle.

Substituting $CM = 0$CM=0, we get the maximum sum:

$$PR + PS = 2PC = 2\sqrt{5}$$PR+PS=2PC=25​

Thus, the maximum value of $(PR + PS)^2$(PR+PS)2 is:

$$(2\sqrt{5})^2 = 4 \times 5 = 20$$(25​)2=4×5=20

Answer: $20$20