## Precalculus (10th Edition)

$x^2 + (y - 290)^2 = 67, 600$
The minimum height of the ferris wheel is equal to: \begin{align*} \text{Maximum height of the wheel} - \text{diameter of the wheel} &= 550 - 520\\ &= 30 \text{ feet} \end{align*} The points $(0, 30)$ and $(0, 550)$ lie on the circle and they are the end points of the vertical diameter of the ferris wheel. Since the center of a circle is the midpoint of a diameter, then the center is at: $\left(\dfrac{0+0}{2}, \dfrac{30+550}{2}\right) = (0, 290)$ The radius $r$ is half of the diameter so $r=\frac{520}{2}=260$. Thus, with $r=260$ and center at $(0, 290)$, the standard equation of the ferris wheel is \begin{align*} (x - 0)^2 + (y - 290)^2 &= 260^2\\ x^2 + (y - 290)^2 &= 67, 600 \end{align*}