Question

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 ________.

Answer 1

Correct Answer: 18

\[ lx+my=1 \] Using homogenization for right angle at origin, \[ 11(l^2+m^2)+6l+8m-2=0 \] Let foot of perpendicular from origin be \((h,k)\). Chord equation: \[ hx+ky=h^2+k^2 \] Comparing with \(lx+my=1\), \[ l=\frac{h}{h^2+k^2},\quad m=\frac{k}{h^2+k^2} \] Substituting, \[ 11+6h+8k-2(h^2+k^2)=0 \] \[ h^2+k^2-3h-4k-\frac{11}{2}=0 \] Replacing \((h,k)\) by \((x,y)\), \[ x^2+y^2-3x-4y-\frac{11}{2}=0 \] Hence \[ \alpha=3,\quad \beta=4,\quad \gamma=\frac{11}{2} \] \[ \alpha+\beta+2\gamma=3+4+11=18 \] \[ \boxed{18} \]