The maximum distance between the point $P(0, 0, 3)$ and the circle $x^2 + y^2 - 2\sqrt{5}x - 4y + 8 = 0; z = 0$ is __________.
Step-by-Step Solution
Key Concept: Maximum distance from external point to a circle equals the distance to center plus the radius.
First, rewrite the circle equation in standard form: $(x - \sqrt{5})^2 + (y - 2)^2 = 5$. This circle lies in the plane $z = 0$ with center $C(\sqrt{5}, 2, 0)$ and radius $r = \sqrt{5}$. The point $P(0, 0, 3)$ is at distance $d = \sqrt{(\sqrt{5})^2 + 2^2 + 3^2} = \sqrt{5 + 4 + 9} = \sqrt{18} = 3\sqrt{2}$ from the center. The maximum distance from $P$ to any point on the circle is $PC + r = 3\sqrt{2} + \sqrt{5}$.
Correct Answer: $3\sqrt{2} + \sqrt{5}$