Find the center and radius of the circle.
-$$x ^ { 2 } + y ^ { 2 } + 10 x - 8 y - 8 = 0$$
A) center: $( - 5,4 )$; radius: 7
B) center: $( 4 , - 5 )$; radius: 7
C) center: $( - 4,5 )$; radius: 49
D) center: $( 5 , - 4 )$; radius: 49

Question

Quizplus · Accepted Answer

To find the center and radius of the circle given by the equation $x^2 + y^2 + 10x - 8y - 8 = 0$, we complete the square for both $x$ and $y$. Rearranging terms: $x^2 + 10x + y^2 - 8y = 8$. Completing the square for $x^2 + 10x$ gives $(x+5)^2 - 25$, and for $y^2 - 8y$ gives $(y-4)^2 - 16$. Substituting these into the original equation and simplifying gives $(x+5)^2 + (y-4)^2 = 49$, indicating a center at $(-5, 4)$ and a radius of $\sqrt{49} = 7$.

Find the Center and Radius of the Circle x2+y2+10x−8y−8=0x ^ { 2 } + y ^ { 2 } + 10 x - 8 y - 8 = 0x2+y2+10x−8y−8=0

Find the Center and Radius of the Circle $x ^ { 2 } + y ^ { 2 } + 10 x - 8 y - 8 = 0$