Let the lines y+2x=√11+7√7 and 2y+x=2√11+6√7 be normal to a circle C:(x-h)^2+(y-k)^2=r^2.

Question

Let the lines $y+2x=\sqrt{11}+7\sqrt7$ and $2y+x=2\sqrt{11}+6\sqrt7$ be normal to a circle $C:(x-h)^2+(y-k)^2=r^2$. If the line $\sqrt{11}y-3x=\dfrac{5\sqrt{77}}3+11$ is tangent to the circle $C$, then the value of $(5h-8k)^2+5r^2$ is equal to:

Answer: 816

Accepted Answer

The centre is the intersection of the two normals: $\left(\dfrac{8\sqrt7}3,\ \sqrt{11}+\dfrac{5\sqrt7}3\right)$.

The radius equals the perpendicular distance from this centre to the given tangent line. Carrying out this computation and substituting into $(5h-8k)^2+5r^2$ gives $816.$

Solutions · 1