Question

determine the equation of the circle with center $(-6,-2)$ containing the point $(3,-6)$. show examples

Explanation:

Step1: Find radius via distance formula

The radius $r$ is the distance between center $(-6,-2)$ and point $(3,-6)$. Use distance formula:
$$r = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2}$$
Substitute $x_1=-6,y_1=-2,x_2=3,y_2=-6$:
$$r = \sqrt{(3-(-6))^2 + (-6-(-2))^2} = \sqrt{(9)^2 + (-4)^2}$$

Step2: Calculate radius squared

Simplify to find $r^2$ (needed for circle equation):
$$r^2 = 9^2 + (-4)^2 = 81 + 16 = 97$$

Step3: Write circle standard equation

Standard form: $(x-h)^2 + (y-k)^2 = r^2$, where $(h,k)=(-6,-2)$:
$$(x-(-6))^2 + (y-(-2))^2 = 97$$

Answer:

$(x+6)^2 + (y+2)^2 = 97$