Question

what is the center of the circle $x^2 + y^2 = 12x + 28$? simplify any fractions.

Explanation:

Step1: Rewrite the equation

We start with the equation of the circle \(x^{2}+y^{2}=12x + 28\). To convert it to the standard form of a circle \((x - h)^{2}+(y - k)^{2}=r^{2}\), we need to complete the square for the \(x\)-terms. First, we move the \(x\)-terms to the left side: \(x^{2}-12x + y^{2}=28\).

Step2: Complete the square for x

For the \(x\)-terms \(x^{2}-12x\), we take half of the coefficient of \(x\) (which is \(- 12\)), so \(\frac{-12}{2}=-6\), and then square it: \((-6)^{2} = 36\). We add and subtract this value (but since we are dealing with an equation, we add it to both sides to keep it balanced). So we have \(x^{2}-12x + 36+y^{2}=28 + 36\).

Step3: Rewrite in standard form

The left side can be written as a perfect square: \((x - 6)^{2}+y^{2}=64\). Now, comparing this with the standard form \((x - h)^{2}+(y - k)^{2}=r^{2}\), where \((h,k)\) is the center of the circle. Here, \(h = 6\) and \(k = 0\) (since there is no \(y\)-term being completed, so the \(y\)-part is \((y-0)^{2}\)).

Answer:

\((6,0)\)