Question

question 18 of 20 what are the coordinates of the center of the circle shown below? express your answer in the form (a,b) without using spaces. $x^{2}+y^{2}-12x - 6y+9 = 0$

Explanation:

Step1: Rewrite the equation in standard form.

Complete the square for \(x\) and \(y\) terms.
For \(x\) - terms: \(x^{2}-12x=(x - 6)^{2}-36\).
For \(y\) - terms: \(y^{2}-6y=(y - 3)^{2}-9\).
The equation \(x^{2}+y^{2}-12x - 6y + 9 = 0\) becomes \((x - 6)^{2}-36+(y - 3)^{2}-9 + 9=0\).

Step2: Simplify the equation.

\((x - 6)^{2}+(y - 3)^{2}=36\).
The standard - form of a circle equation is \((x - a)^{2}+(y - b)^{2}=r^{2}\), where \((a,b)\) is the center of the circle and \(r\) is the radius.

Answer:

(6,3)