Question

find the center and the radius of the following circle.
x² + 12x + y² + 16y = 0

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 (which is 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 equation of the circle

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

$$\begin{align*} x^{2}+12x+y^{2}+16y&=0\\ (x + 6)^{2}-36+(y + 8)^{2}-64&=0\\ (x + 6)^{2}+(y + 8)^{2}&=36 + 64\\ (x + 6)^{2}+(y + 8)^{2}&=100 \end{align*}$$

\]

Step4: Identify the center and radius

The standard form of the equation of a circle is $(x - a)^{2}+(y - b)^{2}=r^{2}$, where $(a,b)$ is the center of the circle and r is the radius.
For the equation $(x + 6)^{2}+(y + 8)^{2}=100=(x-(-6))^{2}+(y - (-8))^{2}$, the center is $(-6,-8)$ and the radius $r = 10$.

Answer:

Center: $(-6,-8)$, Radius: $10$