求穿过点 $(2, 4)$ 和 $(5, 1)$ 的最小圆的面积。答案用 $\\pi$ 表示。 答案 ... $\\text{units}^2$

1. Identify the diameter of the smallest circle and calculate its squared length

For the circle to be the smallest possible, the line segment joining the two points $(2, 4)$ and $(5, 1)$ must be its diameter.
Let $d$ be the diameter.
$d^2 = (5 - 2)^2 + (1 - 4)^2$
$d^2 = 3^2 + (-3)^2 = 9 + 9 = 18$

2. Calculate the area of the circle in terms of $\pi$

The radius $r$ is $\frac{d}{2}$.
Area $= \pi r^2 = \pi \left(\frac{d}{2}\right)^2 = \frac{\pi d^2}{4}$
Area $= \frac{18\pi}{4} = 4.5\pi$