Rewrite the equation in standard form. Find the center and radius of the circle.
-$$x ^ { 2 } + y ^ { 2 } - 12 x - 12 y + 59 = 0$$
A) $( x + 6 ) ^ { 2 } + ( y + 6 ) ^ { 2 } = 13$
center $( - 6 , - 6 ) , r = \sqrt { 13 }$
B) $( x - 6 ) ^ { 2 } + ( y - 6 ) ^ { 2 } = 13$
center $( 6,6 ) , r = \sqrt { 13 }$
C) $( x - 6 ) ^ { 2 } + ( y - 6 ) ^ { 2 } = 13$
center $( - 6 , - 6 ) , r = \sqrt { 13 }$
D) $( x - 6 ) ^ { 2 } + ( y - 6 ) ^ { 2 } = 13$
center $( 6,6 ) , r = 13$

Question

Quizplus · Accepted Answer

The equation can be rewritten by completing the square: $(x - 6)^2 + (y - 6)^2 = 13$. The center is $(6, 6)$ and the radius is $\sqrt{13}$.

Rewrite the Equation in Standard Form x2+y2−12x−12y+59=0x ^ { 2 } + y ^ { 2 } - 12 x - 12 y + 59 = 0x2+y2−12x−12y+59=0

Rewrite the Equation in Standard Form $x ^ { 2 } + y ^ { 2 } - 12 x - 12 y + 59 = 0$