## Precalculus (6th Edition)

$\color{blue}{x^2+y^2=34}$
RECALL: The center-radius form of a circle whose center is at $(h, k)$ and whose radius is $r$ units is: $(x-h)^2+(y-k)^2=r^2$ The given circle has its center at $(0, 0)$ Thus, the tentative equation of the circle is : $(x-0)^2+(y-0)^2=r^2 \\x^2+y^2=r^2$ To find the value of $r^2$, substitute the x and y values of the point $(3, 5)$ to obtain: $3^2+5^2=r^2 \\9+25=r^2 \\34=r^2$ Therefore, the equation of the circle is: $\\\color{blue}{x^2+y^2=34}$