Question

find an equation of the circle that has center (3, -2) and passes through (-5, -5).

Explanation:

Step1: Recall circle equation formula

The standard equation of a circle is $(x-h)^2 + (y-k)^2 = r^2$, where $(h,k)$ is the center and $r$ is the radius.

Step2: Calculate the radius

Use the distance formula between center $(3,-2)$ and point $(-5,-5)$:
$r = \sqrt{(-5-3)^2 + (-5-(-2))^2} = \sqrt{(-8)^2 + (-3)^2} = \sqrt{64 + 9} = \sqrt{73}$
Thus, $r^2 = 73$.

Step3: Substitute center and $r^2$ into formula

Center $(h,k)=(3,-2)$, so substitute into the equation:
$(x-3)^2 + (y+2)^2 = 73$

Answer:

$(x-3)^2 + (y+2)^2 = 73$