## Precalculus (6th Edition)

(a) center-radius: $\color{blue}{(x-3)^2+(y+3)^2=9}$ (b) general form:$\color{red}{x^2+y^2-6x+6y+9=0}$
RECALL: (1) The center-radius form of a circle's equation is $(x-h)^2+(y-k)^2=r^2$ with center at $(h, k)$ and a radius of $r$ units. (2) The general form of a circle's equation is $x^2+y^2+Dx+Ey+F=0$ The center of the circle can be found by finding the the midpoint of $(3, 0)$ and $(3, -6)$. Note that the midpoint of these two points is $(3, -3)$. Thus, the circle has its center at $(3, -3)$. The radius of the circle can be found by finding the distance from the center $(3, -3)$ to $(3, 0)$, which is $3$ units. Thus, the radius of the circle is $3$. (a) The circle has its center at $(3, -3)$ and has a radius of $3$ units. Therefore, the equation of the circle in center-radius form is: $(x-3)^2+(y-(-3))^2=3^2 \\\color{blue}{(x-3)^2+(y+3)^2=9}$ (b) To find the equation of the circle in general form, expand the equation in (a) above then put all terms on the left side to obtain: $(x-3)^2+(y+3)^2=9 \\x^2-6x+9+y^2+6y+9=9 \\x^2+y^2-6x+6y+18=9 \\x^2+y^2-6x+6y+18-9=0 \\\color{red}{x^2+y^2-6x+6y+9=0}$