Answer:
[tex](x+3)^2+(y)^2=3^2[/tex]Step-by-step explanation:
Given the endpoints of the diameter, then the center of the circle is in the middle of the diameter:
The middle point is given as:
[tex]\text{ Mid point= (}\frac{x_1+x_2}{2},\frac{y_1+y_2}{2})[/tex]Therefore, the center of the circle is located at:
[tex]\begin{gathered} \text{Center}=\text{ (}\frac{-6+0}{2},\frac{0+0}{2}) \\ \text{ Center=(-3, 0)} \end{gathered}[/tex]The circle is represented by the following equation:
[tex]\begin{gathered} (x-h)^2+(y-k)^2=r^2 \\ (h,k)\text{ center} \end{gathered}[/tex]By substituting (h,k) and (x,y) (one of the given points), we can solve for r, to find the radius:
[tex]\begin{gathered} (0-(-3))^2+(0-0)^2=r^2 \\ 9=r^2 \\ r=\sqrt[]{9} \\ r=3 \end{gathered}[/tex]Then, this circle is represented by the following equation:
[tex](x+3)^2+(y)^2=3^2[/tex]