Answer :

konrad509
[tex](x-x_1)^2+(y-y_1)^2=r^2\\ (x_1,y_1) - \text{ center}\\\\ r=\dfrac{\text{ diameter}}{2}=\dfrac{10}{2}=5\\\\ (x-(-3))^2+(y-2)^2=5^2\\ \boxed{(x+3)^2+(y-2)^2=25} [/tex]
[tex]The\ equation\ of\ a\ circle:(x-a)^2+(y-b)^2=r^2\\\\(a;\ b)-the\ coordinates\ of\ the\ cener;\ r-the\ radius\\-------------------------------\\The\ center\ (-3;\ 2)\to a=-3;\ b=2\\a\ diameter\ d=2r\ therefore\ r=\frac{1}{2}d\to r=\frac{1}{2}\cdot10=5\\\\Answer:\\\boxed{(x-(-3))^2+(y-2)^2=5^2}\to\boxed{(x+3)^2+(y-2)^2=25}[/tex]