Consider the circle C: x^2+y^2-6x-8y-11=0.

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: 18

Accepted Answer

$C$: centre $(3,4)$, radius $6$. Let $F(h,k)$ be the foot of the perpendicular from $O$ on $AB$; then $AB: hx+ky=h^2+k^2$.

Substituting this relation into $C$'s equation to eliminate $y$ (and separately $x$) gives quadratics in $x_1,x_2$ (roots on $AB\cap C$) and in $y_1,y_2$; using $OA\perp OB$ i.e. $\dfrac{y_1y_2}{x_1x_2}=-1$ and simplifying (using $h^2+k^2\ne0$ since $F\ne O$):

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

So $(x,y)$ satisfies $x^2+y^2-3x-4y-\dfrac{11}{2}=0$: $\alpha=3,\ \beta=4,\ \gamma=\dfrac{11}{2}$.

$$\alpha+\beta+2\gamma=3+4+11=18.$$

Solutions · 1