Answer
$(x-5)^2+(y-6)^2=10$
Work Step by Step
$\bf{\text{Solution Outline:}}$
Use the Center-Radius Form to find the equation of the circle with the following characteristics:
\begin{array}{l}\require{cancel}
\text{center }
(5,6)
,\text{ passing through }
(4,9)
.\end{array}
$\bf{\text{Solution Details:}}$
With the given center, then $h=
5
$ and $k=
6
.$
Using the Center-Radius Form of the equation of circles which is given by $(x-h)^2+(y-k)^2=r^2,$ then the equation of the circle is
\begin{array}{l}\require{cancel}
(x-5)^2+(y-6)^2=r^2
\text{ (*)}
.\end{array}
Since the circle passes through the point $
(4,9)
,$ substituting the coordinates in the equation above results to
\begin{array}{l}\require{cancel}
(4-5)^2+(9-6)^2=r^2
\\\\
(-1)^2+(3)^2=r^2
\\\\
1+9=r^2
\\\\
10=r^2
.\end{array}
Substituting $r^2$ in (*), the equation of the circle is
\begin{array}{l}\require{cancel}
(x-5)^2+(y-6)^2=10
.\end{array}
