JEE Main 2026 · 8 Apr · Shift 2, Mathematics question 24

Chord with foot (h, k). Its equation is $hx+ky=p$, with $p=h^2+k^2$.

Right angle at O. Homogenise the circle with $\tfrac{hx+ky}p=1$; the $x^2$ and $y^2$ coefficients must add to zero:
$$2-\frac{6h+8k}p-\frac{11}p=0\ \Rightarrow\ 2(h^2+k^2)-6h-8k-11=0.$$

Locus. $x^2+y^2-3x-4y-\tfrac{11}2=0$: $\alpha=3$, $\beta=4$, $\gamma=\tfrac{11}2$.

Answer. $\alpha+\beta+2\gamma=18$.