## College Algebra (11th Edition)

$x^2+y^2=34$
$\bf{\text{Solution Outline:}}$ 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}