Question

find the center and radius of the circle.
x^{2}+y^{2}-4x + 16y+64 = 0
center = (?, )
radius =

Explanation:

Step1: Complete the square for x - terms

We have $x^{2}-4x$. Completing the square: $(x - 2)^{2}=x^{2}-4x + 4$, so $x^{2}-4x=(x - 2)^{2}-4$.

Step2: Complete the square for y - terms

We have $y^{2}+16y$. Completing the square: $(y + 8)^{2}=y^{2}+16y+64$, so $y^{2}+16y=(y + 8)^{2}-64$.

Step3: Rewrite the circle equation

Substitute the completed - square expressions into the original equation $x^{2}+y^{2}-4x + 16y+64 = 0$.
\[

$$\begin{align*} (x - 2)^{2}-4+(y + 8)^{2}-64+64&=0\\ (x - 2)^{2}+(y + 8)^{2}&=4 \end{align*}$$

\]

Step4: Identify the center and radius

The standard form of a circle equation is $(x - a)^{2}+(y - b)^{2}=r^{2}$, where $(a,b)$ is the center and $r$ is the radius.
For the equation $(x - 2)^{2}+(y + 8)^{2}=4=(x - 2)^{2}+(y-(-8))^{2}=2^{2}$, the center is $(2,-8)$ and the radius $r = 2$.

Answer:

Center = $(2,-8)$
Radius = $2$