Question

question
determine the equation of the circle with center $(-6, -2)$ containing the point $(3, -6)$.
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Explanation:

Step1: Calculate the radius (distance)

The distance formula between two points $(x_1,y_1)$ and $(x_2,y_2)$ is $r = \sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$. Substitute $(x_1,y_1)=(-6,-2)$ and $(x_2,y_2)=(3,-6)$:
$r = \sqrt{(3-(-6))^2+(-6-(-2))^2} = \sqrt{(9)^2+(-4)^2} = \sqrt{81+16} = \sqrt{97}$

Step2: Find $r^2$

Square the radius to use in the circle equation:
$r^2 = (\sqrt{97})^2 = 97$

Step3: Write circle equation

The standard circle equation is $(x-h)^2+(y-k)^2=r^2$, where $(h,k)=(-6,-2)$:
$(x-(-6))^2+(y-(-2))^2=97$

Answer:

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