JEE Main 2026 (Online) 8th April Evening ShiftQuestion 1
Consider the circle $C:x^{2}+y^{2}-6x-8y-11=0$. Let a variable chord $AB$ of the circle $C$ subtend a right angle at the origin. If the locus of the foot of the perpendicular drawn from the origin on the chord $AB$ is the circle $x^{2}+y^{2}-\alpha x-\beta y-\gamma=0$, then $\alpha+\beta+2\gamma$ is equal to $\_\_\_\_\_\_$ .
Circle
Answer$18$
Solution
Let foot be $(h,k)$; chord: $hx+ky=h^2+k^2=m$. Homogenise the circle: $x^2+y^2-(6x+8y)\frac{hx+ky}{m}-11\left(\frac{hx+ky}{m}\right)^2=0$. Right angle at origin means coefficient of $x^2$ + coefficient of $y^2$ $=0$. This gives $2(h^2+k^2)-6h-8k-11=0$, i.e. $x^2+y^2-3x-4y-\frac{11}2=0$. $\alpha=3,\beta=4,\gamma=\frac{11}2$, so $\alpha+\beta+2\gamma=18$.