Question

the equation of a circle is given below. identify the radius and the center. then graph the circle.
$x^{2}+y^{2}-4x - 6y-12 = 0$
radius:
center:

Explanation:

Step1: Complete the square for x - terms

\[

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

\]

Step2: Complete the square for y - terms

\[

$$\begin{align*} y^{2}-6y&=(y - 3)^{2}-9 \end{align*}$$

\]

Step3: Rewrite the circle equation

\[

$$\begin{align*} x^{2}+y^{2}-4x - 6y-12&=0\\ (x - 2)^{2}-4+(y - 3)^{2}-9-12&=0\\ (x - 2)^{2}+(y - 3)^{2}&=4 + 9+12\\ (x - 2)^{2}+(y - 3)^{2}&=25 \end{align*}$$

\]

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 \((x - 2)^{2}+(y - 3)^{2}=25\), the center is \((2,3)\) and the radius \(r = 5\).

Answer:

Radius: 5
Center: (2,3)