Answer
$\text{Center: }
(3,5)
\\\text{Radius: }
2$
Work Step by Step
$\bf{\text{Solution Outline:}}$
To find the Center-Radius Form of the given equation, $
x^2+y^2-6x-10y+30=0
,$ complete the square for both $x$ and $y$ variables.
$\bf{\text{Solution Details:}}$
Grouping the $x$-variables and the $y$-variables and transposing the constant, the given equation is equivalent to
\begin{array}{l}\require{cancel}
(x^2-6x)+(y^2-10y)=-30
.\end{array}
Since the coefficient of $x$ is $
-6
,$ add $\left(\dfrac{
-6
}{2}\right)^2=
9
$ on both sides of the equation to complete the square for $x$. Hence, the equation above is equivalent to
\begin{array}{l}\require{cancel}
(x^2-6x+9)+(y^2-10y)=-30+9
\\\\
(x-3)^2+(y^2-10y)=-21
.\end{array}
Since the coefficient of $y$ is $
-10
,$ add $\left(\dfrac{
-10
}{2}\right)^2=
25
$ on both sides of the equation to complete the square for $y$. Hence, the equation above is equivalent to
\begin{array}{l}\require{cancel}
(x-3)^2+(y^2-10y+25)=-21+25
\\\\
(x-3)^2+(y-5)^2=4
.\end{array}
Using the Center-Radius Form of the equation of circles which is given by $(x-h)^2+(y-k)^2=r^2,$ where the center is $(h,k)$ and the radius is $r,$ the equation above is equivalent to
\begin{array}{l}\require{cancel}
(x-(3))^2+(y-(5))^2=2^2
.\end{array}
Hence the equation of the circle above has the following characteristics:
\begin{array}{l}\require{cancel}
\text{Center: }
(3,5)
\\\text{Radius: }
2
.\end{array}
