Question

what is the center of the circle ( x^2 + y^2 - 16x = 0 )? simplify any fractions.

Explanation:

Step1: Recall the standard circle equation

The standard form of a circle's equation is \((x - h)^2 + (y - k)^2 = r^2\), where \((h, k)\) is the center and \(r\) is the radius. We need to rewrite the given equation \(x^2 + y^2 - 16x = 0\) into this form by completing the square for the \(x\)-terms.

Step2: Group and complete the square for \(x\)

Group the \(x\)-terms: \((x^2 - 16x) + y^2 = 0\). To complete the square for \(x^2 - 16x\), we take half of \(-16\) (which is \(-8\)), square it (\((-8)^2 = 64\)), and add it to both sides:
\[
(x^2 - 16x + 64) + y^2 = 64
\]
This factors to:
\[
(x - 8)^2 + y^2 = 64
\]

Step3: Identify the center

Comparing with the standard form \((x - h)^2 + (y - k)^2 = r^2\), we see that \(h = 8\) and \(k = 0\).

Answer:

\((8, 0)\)