Find the center and radius of the circle.
-$$x ^ { 2 } + y ^ { 2 } - 12 x + 2 y - 44 = 0$$
A) center: $( 6 , - 1 )$; radius: 9
B) center: $( 1 , - 6 )$; radius: 81
C) center: $( - 6,1 )$; radius: 81
D) center: $( - 1,6 )$; radius: 9

Question

Quizplus · Accepted Answer

To find the center and radius of the circle given by the equation $x^2 + y^2 - 12x + 2y - 44 = 0$, we complete the squares for $x$ and $y$. The equation can be rewritten as:$$x^2 - 12x + y^2 + 2y = 44$$Completing the square for $x^2 - 12x$, we add and subtract $(\frac{12}{2})^2 = 36$, and for $y^2 + 2y$, we add and subtract $(\frac{2}{2})^2 = 1$, resulting in:$$x^2 - 12x + 36 + y^2 + 2y + 1 = 44 + 36 + 1$$$$x^2 - 12x + 36 + y^2 + 2y + 1 = 81$$$$(x - 6)^2 + (y + 1)^2 = 81$$This equation is in the standard form of a circle $(x - h)^2 + (y - k)^2 = r^2$, where $(h, k)$ is the center and $r$ is the radius. Therefore, the center is $(6, -1)$ and the radius is $\sqrt{81} = 9$.

Find the Center and Radius of the Circle x2+y2−12x+2y−44=0x ^ { 2 } + y ^ { 2 } - 12 x + 2 y - 44 = 0x2+y2−12x+2y−44=0

Find the Center and Radius of the Circle $x ^ { 2 } + y ^ { 2 } - 12 x + 2 y - 44 = 0$