Question

find the center and the radius of the following circle.
x² - 12x + y² + 16y = 0
the center is
(type an ordered pair.)
the radius is
(simplify your answer.)
use the graphing tool to graph the circle.
click to enlarge graph

Explanation:

Step1: Complete the square for x - terms

We have $x^{2}-12x$. To complete the square, we take half of the coefficient of $x$ (-12), square it. Half of - 12 is -6, and $(-6)^{2}=36$. So $x^{2}-12x=(x - 6)^{2}-36$.

Step2: Complete the square for y - terms

We have $y^{2}+16y$. Half of 16 is 8, and $8^{2}=64$. So $y^{2}+16y=(y + 8)^{2}-64$.

Step3: Rewrite the circle equation

The original equation $x^{2}-12x + y^{2}+16y=0$ becomes $(x - 6)^{2}-36+(y + 8)^{2}-64 = 0$.
Rearranging, we get $(x - 6)^{2}+(y + 8)^{2}=36 + 64$.

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 $(x - 6)^{2}+(y + 8)^{2}=100$, the center is $(6,-8)$ and the radius $r=\sqrt{100}=10$.

Answer:

The center is $(6,-8)$
The radius is $10$