Find the center and radius of the circle.
-$$2 x ^ { 2 } + 2 y ^ { 2 } - 16 x - 4 y + 16 = 0$$
A) center: $( 4,1 )$, radius: 3
B) center: $( 1,4 )$, radius: 3
C) center: $( - 4 , - 1 )$, radius: 9
D) center: $( - 1 , - 4 )$, radius: 9

Question

Quizplus · Accepted Answer

To find the center and radius of the circle given by the equation $2x^2 + 2y^2 - 16x - 4y + 16 = 0$, we first complete the square for both $x$ and $y$. Divide the equation by 2 to simplify: $x^2 + y^2 - 8x - 2y + 8 = 0$. For $x$, add and subtract $(8/2)^2 = 16$, and for $y$, add and subtract $(2/2)^2 = 1$, resulting in $x^2 - 8x + 16 + y^2 - 2y + 1 = -8 + 16 + 1$. This simplifies to $(x - 4)^2 + (y - 1)^2 = 9$, indicating a center at $(4, 1)$ and a radius of $\sqrt{9} = 3$.

Find the Center and Radius of the Circle 2x2+2y2−16x−4y+16=02 x ^ { 2 } + 2 y ^ { 2 } - 16 x - 4 y + 16 = 02x2+2y2−16x−4y+16=0

Find the Center and Radius of the Circle $2 x ^ { 2 } + 2 y ^ { 2 } - 16 x - 4 y + 16 = 0$