Find the center and radius of the circle whose equation is given.
-$x^{2}+y^{2}+12 x+8 y+52=81$
A) center: $(4,6)$; radius: 81
B) center: $(6,4)$; radius: 81
C) center: $(-6,-4)$; radius: 9
D) center: $(-4,-6)$; radius: 9

Question

Quizplus · Accepted Answer

To find the center and radius of the circle given by the equation $x^{2}+y^{2}+12x+8y+52=81$, we complete the squares for $x$ and $y$. The equation can be rewritten as $(x^2 + 12x) + (y^2 + 8y) = 29$. Completing the square for $x^2 + 12x$ gives $(x+6)^2 - 36$ and for $y^2 + 8y$ gives $(y+4)^2 - 16$. Adding these to both sides of the equation, we get $(x+6)^2 + (y+4)^2 = 29 + 36 + 16 = 81$. Thus, the center is $(-6, -4)$ and the radius is $\sqrt{81} = 9$.

Find the Center and Radius of the Circle Whose Equation x2+y2+12x+8y+52=81x^{2}+y^{2}+12 x+8 y+52=81x2+y2+12x+8y+52=81

Find the Center and Radius of the Circle Whose Equation $x^{2}+y^{2}+12 x+8 y+52=81$