JEE Main 2026 — Mathematics Coordinate Geometry

Let the foot of the perpendicular from the origin to the chord $AB$AB be $P(h, k)$P(h,k).

The slope of $OP$OP is $\dfrac{k}{h}$hk​. Since the chord $AB$AB is perpendicular to $OP$OP, its equation is given by:

$$y - k = -\dfrac{h}{k}(x - h) \Rightarrow hx + ky = h^2 + k^2$$y−k=−kh​(x−h)⇒hx+ky=h2+k2

This can be rewritten as $\dfrac{hx + ky}{h^2 + k^2} = 1$h2+k2hx+ky​=1.

The chord $AB$AB subtends a right angle at the origin. We homogenize the equation of the circle $x^2 + y^2 - 6x - 8y - 11 = 0$x2+y2−6x−8y−11=0 with the equation of the chord to find the joint equation of the lines $OA$OA and $OB$OB:

$$x^2 + y^2 - (6x + 8y)\left(\dfrac{hx + ky}{h^2 + k^2}\right) - 11\left(\dfrac{hx + ky}{h^2 + k^2}\right)^2 = 0$$x2+y2−(6x+8y)(h2+k2hx+ky​)−11(h2+k2hx+ky​)2=0

Since the lines $OA$OA and $OB$OB are perpendicular, the sum of the coefficients of $x^2$x2 and $y^2$y2 in this homogenized equation must be zero.

Coefficient of $x^2$x2: $1 - \dfrac{6h}{h^2 + k^2} - \dfrac{11h^2}{(h^2 + k^2)^2}$1−h2+k26h​−(h2+k2)211h2​

Coefficient of $y^2$y2: $1 - \dfrac{8k}{h^2 + k^2} - \dfrac{11k^2}{(h^2 + k^2)^2}$1−h2+k28k​−(h2+k2)211k2​

Equating the sum of these coefficients to zero:

$$2 - \dfrac{6h + 8k}{h^2 + k^2} - \dfrac{11(h^2 + k^2)}{(h^2 + k^2)^2} = 0$$2−h2+k26h+8k​−(h2+k2)211(h2+k2)​=0

$$2 - \dfrac{6h + 8k}{h^2 + k^2} - \dfrac{11}{h^2 + k^2} = 0$$2−h2+k26h+8k​−h2+k211​=0

$$2(h^2 + k^2) - 6h - 8k - 11 = 0$$2(h2+k2)−6h−8k−11=0

$$h^2 + k^2 - 3h - 4k - \dfrac{11}{2} = 0$$h2+k2−3h−4k−211​=0

Replacing $(h, k)$(h,k) with $(x, y)$(x,y), the locus of the foot of the perpendicular is:

$$x^2 + y^2 - 3x - 4y - \dfrac{11}{2} = 0$$x2+y2−3x−4y−211​=0

Comparing this with the given locus equation $x^2 + y^2 - \alpha x - \beta y - \gamma = 0$x2+y2−αx−βy−γ=0, we get:

$\alpha = 3$α=3, $\beta = 4$β=4, $\gamma = \dfrac{11}{2}$γ=211​

Therefore, the value of $\alpha + \beta + 2\gamma$α+β+2γ is:

$$3 + 4 + 2\left(\dfrac{11}{2}\right) = 7 + 11 = 18$$3+4+2(211​)=7+11=18

Answer: $18$18