Let P be a moving point on the circle x^2+y^2-6x-8y+21=0.

Question

Let $P$ be a moving point on the circle $x^2+y^2-6x-8y+21=0$. Then the maximum distance of $P$ from the vertex of the parabola $y=-(x+3)^2-4$ is equal to:

Accepted Answer

Vertex of the parabola: $(-3,-4)$. Circle: $(x-3)^2+(y-4)^2=4$, centre $(3,4)$, radius $2$.

$$VC=\sqrt{(3+3)^2+(4+4)^2}=\sqrt{36+64}=10.$$

Maximum distance $=VC+r=10+2=12$.

Solutions · 1