Find the center and radius of the circle.
-$$x ^ { 2 } + y ^ { 2 } + 4 x - 16 y - 13 = 0$$
A) center: $( 2 , - 8 )$; radius: 81
B) center: $( - 2,8 )$; radius: 9
C) center: $( - 8,2 )$; radius: 81
D) center: $( 8 , - 2 )$; radius: 9

Question

Quizplus · Accepted Answer

To find the center and radius of the circle given by the equation $x^2 + y^2 + 4x - 16y - 13 = 0$, we complete the square for both $x$ and $y$. Rearranging terms: $(x^2 + 4x) + (y^2 - 16y) = 13$. Completing the square for $x^2 + 4x$ gives $(x + 2)^2 - 4$, and for $y^2 - 16y$ gives $(y - 8)^2 - 64$. Substituting these into the equation and simplifying gives $(x + 2)^2 + (y - 8)^2 = 81$, which shows the center is $(-2, 8)$ and the radius is $\sqrt{81} = 9$.

Find the Center and Radius of the Circle x2+y2+4x−16y−13=0x ^ { 2 } + y ^ { 2 } + 4 x - 16 y - 13 = 0x2+y2+4x−16y−13=0

Find the Center and Radius of the Circle $x ^ { 2 } + y ^ { 2 } + 4 x - 16 y - 13 = 0$