JEE Main 2026 — Circle Question with Solution

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

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

$y-k=-\frac{h}{k}(x-h)\Rightarrow hx+ky=h^2+k^2$

This can be rewritten as $\frac{hx+ky}{h^2+k^2}=1$.

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

$x^2+y^2-(6x+8y)\left(\frac{hx+ky}{h^2+k^2}\right)-11{\left(\frac{hx+ky}{h^2+k^2}\right)}^2=0$

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

Coefficient of $x^2$: $1-\frac{6h}{h^2+k^2}-\frac{11h^2}{(h^2+k^2{)}^2}$

Coefficient of $y^2$: $1-\frac{8k}{h^2+k^2}-\frac{11k^2}{(h^2+k^2{)}^2}$

Equating the sum of these coefficients to zero:

$2-\frac{6h+8k}{h^2+k^2}-\frac{11(h^2+k^2)}{(h^2+k^2{)}^2}=0$

$2-\frac{6h+8k}{h^2+k^2}-\frac{11}{h^2+k^2}=0$

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

$h^2+k^2-3h-4k-\frac{11}{2}=0$

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

$x^2+y^2-3x-4y-\frac{11}{2}=0$

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

$\alpha =3$, $\beta =4$, $\gamma =\frac{11}{2}$

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

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

Answer: $18$