Question

write the equation in standard form for the circle $x^{2}+y^{2}-16x + 45 = 0$.

Explanation:

Step1: Group x-terms and move constant

Group the \(x\)-terms and move the constant to the right: \(x^{2}-16x + y^{2}=-45\).

Step2: Complete the square for x

For \(x^{2}-16x\), take half of \(-16\) (which is \(-8\)), square it (\(64\)), and add to both sides: \(x^{2}-16x + 64 + y^{2}=-45 + 64\).

Step3: Rewrite as perfect squares

Rewrite \(x^{2}-16x + 64\) as \((x - 8)^{2}\) and simplify the right side: \((x - 8)^{2}+y^{2}=19\).

Answer:

\((x - 8)^{2}+y^{2}=19\)