Find the center and radius of the circle.
-$2 x ^ { 2 } + 2 y ^ { 2 } + 8 x + 12 y + 24 = 0$
A) $( - 3 , - 2 ) , \mathrm { r } = 1$
B) $( - 2 , - 3 ) , \mathrm { r } = 1$
C) $( 2,3 ) , r = 1$
D) $( 3,2 ) , \mathrm { r } = 1$

Question

Quizplus · Accepted Answer

To find the center and radius, first divide the equation by 2: $x^2 + y^2 + 4x + 6y + 12 = 0$. Complete the square: $(x+2)^2 - 4 + (y+3)^2 - 9 + 12 = 0$. Simplify to $(x+2)^2 + (y+3)^2 = 1$. The center is $(-2, -3)$ and the radius is 1.

Find the Center and Radius of the Circle 2x2+2y2+8x+12y+24=02 x ^ { 2 } + 2 y ^ { 2 } + 8 x + 12 y + 24 = 02x2+2y2+8x+12y+24=0

Find the Center and Radius of the Circle $2 x ^ { 2 } + 2 y ^ { 2 } + 8 x + 12 y + 24 = 0$