Question

find the center and the radius of the following circle. ( x^2 + 12x + y^2 - 16y = 0 ) the center is (square). (type an ordered pair.)

Explanation:

Step1: Complete the square for x-terms

For the \(x\)-terms \(x^{2}+12x\), we use the formula \((a + b)^2=a^{2}+2ab + b^{2}\). Here, \(2ab = 12x\) and \(a=x\), so \(b = 6\) (since \(2\times x\times6=12x\)). Then \(x^{2}+12x=(x + 6)^{2}-36\).

Step2: Complete the square for y-terms

For the \(y\)-terms \(y^{2}-16y\), using the same formula, \(2ab=-16y\) with \(a = y\), so \(b=-8\) (since \(2\times y\times(-8)=-16y\)). Then \(y^{2}-16y=(y - 8)^{2}-64\).

Step3: Rewrite the circle equation

Substitute the completed squares back into the original equation \(x^{2}+12x+y^{2}-16y = 0\):
\[

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

\]
The standard form of a circle is \((x - h)^{2}+(y - k)^{2}=r^{2}\), where \((h,k)\) is the center and \(r\) is the radius. Comparing \((x + 6)^{2}+(y - 8)^{2}=10^{2}\) with the standard form, we get \(h=-6\), \(k = 8\), and \(r = 10\).

Answer:

The center is \((-6,8)\) and the radius is \(10\). For the center (as asked in the box), the answer is \((-6,8)\).