Find the center and radius of the circle.
-$$x ^ { 2 } + y ^ { 2 } + 16 x + 6 y + 24 = 0$$
A) center: $( - 3 , - 8 )$; radius: 7
B) center: $( - 8 , - 3 )$; radius: 7
C) center: $( 3,8 )$; radius: 49
D) center: $( 8,3 )$; radius: 49

Question

Quizplus · Accepted Answer

To find the center and radius of the circle given by the equation $x^2 + y^2 + 16x + 6y + 24 = 0$, we complete the square for both $x$ and $y$. For $x$, we add and subtract $(\frac{16}{2})^2 = 64$, and for $y$, we add and subtract $(\frac{6}{2})^2 = 9$. This gives us $x^2 + 16x + 64 + y^2 + 6y + 9 = -24 + 64 + 9$, simplifying to $(x+8)^2 + (y+3)^2 = 49$. Thus, the center is $(-8, -3)$ and the radius is $\sqrt{49} = 7$.

Find the Center and Radius of the Circle x2+y2+16x+6y+24=0x ^ { 2 } + y ^ { 2 } + 16 x + 6 y + 24 = 0x2+y2+16x+6y+24=0

Find the Center and Radius of the Circle $x ^ { 2 } + y ^ { 2 } + 16 x + 6 y + 24 = 0$