Question

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

Explanation:

Step1: Complete the square for x-terms

For the \(x\)-terms \(x^{2}+16x\), we use the formula \((a + b)^2=a^{2}+2ab + b^{2}\). Here, \(2ab = 16x\) and \(a=x\), so \(b = 8\) (since \(2\times x\times8=16x\)). Then \(x^{2}+16x=(x + 8)^{2}-64\).

Step2: Complete the square for y-terms

For the \(y\)-terms \(y^{2}+12y\), using the same formula, \(2ab = 12y\) with \(a = y\), so \(b = 6\) (since \(2\times y\times6 = 12y\)). Then \(y^{2}+12y=(y + 6)^{2}-36\).

Step3: Rewrite the circle equation

Substitute the completed squares back into the original equation:
\[

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

\]
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 \((x + 8)^{2}+(y + 6)^{2}=100\) with the standard form, we have \(h=-8\), \(k = - 6\), and \(r^{2}=100\) (so \(r = 10\)).

Answer:

The center is \((-8,-6)\) and the radius is \(10\).