Find the equation of the circle that has a diameter with the given endpoints.
-$(-6,6)$ and $(0,-2)$
A) $(x-3)^{2}+(y+2)^{2}=25$
B) $(x+3)^{2}+(y-2)^{2}=25$
C) $(x+3)^{2}+(y-2)^{2}=5$
D) $(x-3)^{2}+(y+2)^{2}=100$

Question

Quizplus · Accepted Answer

The center of the circle is the midpoint of the diameter, which is $\left(\frac{-6+0}{2}, \frac{6-2}{2}\right) = (-3,2)$. The radius is half the distance between the endpoints, calculated using the distance formula: $\sqrt{(-6-0)^2 + (6+2)^2} = \sqrt{36 + 64} = \sqrt{100} = 10$. Therefore, the radius squared is $10^2 = 100$. The equation of the circle is $(x+3)^{2}+(y-2)^{2}=100$.

Find the Equation of the Circle That Has a Diameter (−6,6)(-6,6)(−6,6)

Find the Equation of the Circle That Has a Diameter $(-6,6)$