Question

find the center and radius of the circle.
$x^{2}+y^{2}-6x - 12y-8 = 0$
the center is
(type an ordered pair.)

Explanation:

Step1: Complete the square for x - terms

Group the x - terms and y - terms: $(x^{2}-6x)+(y^{2}-12y)=8$. For the x - terms, $x^{2}-6x=(x - 3)^{2}-9$ (using $(a - b)^2=a^{2}-2ab + b^{2}$, here $a = x$ and $b = 3$).

Step2: Complete the square for y - terms

For the y - terms, $y^{2}-12y=(y - 6)^{2}-36$ (where $a = y$ and $b = 6$). So the equation becomes $(x - 3)^{2}-9+(y - 6)^{2}-36=8$.

Step3: Rewrite in standard form

Rearrange the equation to get the standard form of the circle equation $(x - h)^{2}+(y - k)^{2}=r^{2}$. $(x - 3)^{2}+(y - 6)^{2}=8 + 9+36$. So $(x - 3)^{2}+(y - 6)^{2}=53$.

Answer:

Center: $(3,6)$; Radius: $\sqrt{53}$