Question

what is the center of the circle ( x^2 + y^2 - 20y - 21 = 0 )? simplify any fractions.

Explanation:

Step1: Recall circle equation form

The standard form of a circle's equation is \((x - h)^2 + (y - k)^2 = r^2\), where \((h, k)\) is the center. We need to rewrite the given equation \(x^2 + y^2 - 20y - 21 = 0\) into this form by completing the square for the \(y\)-terms (and \(x\)-terms, but here \(x\) has no linear term).

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

Group the \(y\)-terms: \(x^2 + (y^2 - 20y) - 21 = 0\). To complete the square for \(y^2 - 20y\), take half of \(-20\) (which is \(-10\)), square it (\((-10)^2 = 100\)), and add and subtract it (but since we add it inside the parentheses, we must add it to the other side to balance the equation). So:

\(x^2 + (y^2 - 20y + 100 - 100) - 21 = 0\)

Rewrite as: \(x^2 + (y^2 - 20y + 100) - 100 - 21 = 0\)

Step3: Rewrite in standard form

Now, \(y^2 - 20y + 100 = (y - 10)^2\), so the equation becomes:

\(x^2 + (y - 10)^2 - 121 = 0\)

Then, \((x - 0)^2 + (y - 10)^2 = 11^2\)

Answer:

\((0, 10)\)