Find the center and radius of the circle.
-$x^{2}+y^{2}+8 x+12 y+16=0$
A) $(4,6) ; r=36$
B) $(6,4) ; r=36$
C) $(-4,-6) ; \mathrm{r}=6$
D) $(-6,-4) ; r=6$

Question

Quizplus · Accepted Answer

To find the center and radius of the circle given by the equation $x^{2}+y^{2}+8 x+12 y+16=0$, we complete the squares for $x$ and $y$. The equation can be rewritten as $(x^2 + 8x) + (y^2 + 12y) = -16$. Completing the square for $x^2 + 8x$ gives $(x+4)^2 - 16$, and for $y^2 + 12y$ gives $(y+6)^2 - 36$. So, the equation becomes $(x+4)^2 + (y+6)^2 = 36$. This shows the center is at $(-4, -6)$ and the radius $r = \sqrt{36} = 6$.

Find the Center and Radius of the Circle x2+y2+8x+12y+16=0x^{2}+y^{2}+8 x+12 y+16=0x2+y2+8x+12y+16=0

Find the Center and Radius of the Circle $x^{2}+y^{2}+8 x+12 y+16=0$