Rewrite the equation in standard form. Find the center and radius of the circle.
-$$x ^ { 2 } + y ^ { 2 } - 12 y + 20 = 0$$
A) $x ^ { 2 } + ( y + 6 ) ^ { 2 } = 16$
center $( 0 , - 6 ) , \mathrm { r } = 4$
B) $x ^ { 2 } + ( y - 6 ) ^ { 2 } = 16$
center $( 0,6 ) , \mathrm { r } = 4$
C) $( x - 6 ) ^ { 2 } + y ^ { 2 } = 16$
center $( 6,0 ) , \mathrm { r } = 4$
D) $x ^ { 2 } + ( y - 6 ) ^ { 2 } = 16$
center $( 0 , - 6 ) , \mathrm { r } = 4$

Question

Quizplus · Accepted Answer

The equation can be rewritten by completing the square for the y terms: $x^2 + (y^2 - 12y) + 20 = 0$. Completing the square gives $x^2 + (y - 6)^2 = 16$, which is in standard form. The center is $(0, 6)$ and the radius is 4.

Rewrite the Equation in Standard Form x2+y2−12y+20=0x ^ { 2 } + y ^ { 2 } - 12 y + 20 = 0x2+y2−12y+20=0

Rewrite the Equation in Standard Form $x ^ { 2 } + y ^ { 2 } - 12 y + 20 = 0$