Question

find the center and radius of the circle. 4x^2 + 4y^2 - 16x + 32y + 16 = 0 the center is . (type an ordered pair.)

Explanation:

Step1: Divide the equation by 4

Divide the given equation $4x^{2}+4y^{2}-16x + 32y+16 = 0$ by 4 to simplify. We get $x^{2}+y^{2}-4x + 8y+4=0$.

Step2: Complete the square for x - terms

For the $x$ - terms in $x^{2}-4x$, we have $(x - 2)^{2}=x^{2}-4x + 4$. So $x^{2}-4x=(x - 2)^{2}-4$.

Step3: Complete the square for y - terms

For the $y$ - terms in $y^{2}+8y$, we have $(y + 4)^{2}=y^{2}+8y+16$. So $y^{2}+8y=(y + 4)^{2}-16$.

Step4: Rewrite the equation

Substitute the completed - square expressions into the simplified equation:
\[

$$\begin{align*} (x - 2)^{2}-4+(y + 4)^{2}-16+4&=0\\ (x - 2)^{2}+(y + 4)^{2}&=16 \end{align*}$$

\]
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.

Answer:

The center is $(2,-4)$ and the radius is 4.