Complete the square and write the equation in standard form. Then give the center and radius of the circle.
-$$x ^ { 2 } + y ^ { 2 } - 12 x - 6 y + 33 = 0$$
A) $( x - 6 ) ^ { 2 } + ( y - 3 ) ^ { 2 } = 12$
$( 6,3 ) , r = 2 \sqrt { 3 }$
B) $( x - 6 ) ^ { 2 } + ( y - 3 ) ^ { 2 } = 12$
$( 6,3 ) , r = 12$
C) $( x - 6 ) ^ { 2 } + ( y - 3 ) ^ { 2 } = 12$
$( - 6 , - 3 ) , r = 2 \sqrt { 3 }$
D) $( x + 6 ) ^ { 2 } + ( y + 3 ) ^ { 2 } = 12$
$( - 6 , - 3 ) , \mathrm { r } = 2 \sqrt { 3 }$ Graph the equation.

Question

Quizplus · Accepted Answer

To complete the square, the equation $x^2 + y^2 - 12x - 6y + 33 = 0$ is rearranged as $(x^2 - 12x) + (y^2 - 6y) = -33$. Completing the square for each binomial, we add $(\frac{12}{2})^2 = 36$ for $x$ and $(\frac{6}{2})^2 = 9$ for $y$ to both sides, resulting in $(x^2 - 12x + 36) + (y^2 - 6y + 9) = 12$, which simplifies to $(x - 6)^2 + (y - 3)^2 = 12$. The center of the circle is at $(6, 3)$, and the radius $r$ is the square root of 12, which is $2\sqrt{3}$.

Complete the Square and Write the Equation in Standard Form x2+y2−12x−6y+33=0x ^ { 2 } + y ^ { 2 } - 12 x - 6 y + 33 = 0x2+y2−12x−6y+33=0

Complete the Square and Write the Equation in Standard Form $x ^ { 2 } + y ^ { 2 } - 12 x - 6 y + 33 = 0$