JEE Main 2026 — Mathematics Coordinate Geometry

The given circle $C: (x-4)^2 + (y+3)^2 = 9$C:(x−4)2+(y+3)2=9 has its center at $C_0(4, -3)$C0​(4,−3) and radius $r = 3$r=3.

The line $QR$QR has the equation $x - y = 4$x−y=4, which has a slope of $1$1.

Since $PQ = PR$PQ=PR, the point $P(\alpha, \beta)$P(α,β) must lie on the perpendicular bisector of the chord $QR$QR. The perpendicular bisector of any chord of a circle passes through its center. Therefore, the perpendicular bisector passes through $C_0(4, -3)$C0​(4,−3) and has a slope of $-1$−1 (since it is perpendicular to $QR$QR).

The equation of the perpendicular bisector is:

$$y - (-3) = -1(x - 4)$$y−(−3)=−1(x−4)

$$x + y = 1$$x+y=1

Since $P(\alpha, \beta)$P(α,β) lies on this perpendicular bisector, we have:

$$\alpha + \beta = 1 \Rightarrow \beta = 1 - \alpha$$α+β=1⇒β=1−α

Also, $P(\alpha, \beta)$P(α,β) lies on the circle $C$C, so its coordinates must satisfy the circle's equation:

$$(\alpha - 4)^2 + (\beta + 3)^2 = 9$$(α−4)2+(β+3)2=9

Substituting $\beta = 1 - \alpha$β=1−α into the equation:

$$(\alpha - 4)^2 + (1 - \alpha + 3)^2 = 9$$(α−4)2+(1−α+3)2=9

$$(\alpha - 4)^2 + (4 - \alpha)^2 = 9$$(α−4)2+(4−α)2=9

$$2(\alpha - 4)^2 = 9$$2(α−4)2=9

$$\alpha - 4 = \pm \dfrac{3}{\sqrt{2}}$$α−4=±2​3​

$$\alpha = 4 \pm \dfrac{3}{\sqrt{2}}$$α=4±2​3​

We need to find the value of $(6\alpha + 8\beta)^2$(6α+8β)2. Substituting $\beta = 1 - \alpha$β=1−α:

$$6\alpha + 8\beta = 6\alpha + 8(1 - \alpha) = 8 - 2\alpha$$6α+8β=6α+8(1−α)=8−2α

Now, substituting the value of $\alpha$α:

$$8 - 2\alpha = 8 - 2\left(4 \pm \dfrac{3}{\sqrt{2}}\right) = \mp \dfrac{6}{\sqrt{2}} = \mp 3\sqrt{2}$$8−2α=8−2(4±2​3​)=∓2​6​=∓32​

Squaring this value gives:

$$(6\alpha + 8\beta)^2 = (\mp 3\sqrt{2})^2 = 18$$(6α+8β)2=(∓32​)2=18

Answer: $18$18