#### Answer

$x^2+y^2=34$

#### 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 }
(0,0)
,\text{ passing through }
(3,5)
.\end{array}
$\bf{\text{Solution Details:}}$
With the given center, then $h=
0
$ and $k=
0
.$
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-0)^2+(y-0)^2=r^2
\\\\
x^2+y^2=r^2
\text{ (*)}
.\end{array}
Since the circle passes through the point $
(3,5)
,$ substituting the coordinates in the equation above results to
\begin{array}{l}\require{cancel}
3^2+5^2=r^2
\\\\
9+25=r^2
\\\\
34=r^2
.\end{array}
Substituting $r^2$ in (*), the equation of the circle is
\begin{array}{l}\require{cancel}
x^2+y^2=34
.\end{array}