QuestionJuly 22, 2026

Find an equation of the circle whose diameter has endpoints (-6,-3) and (-2,3) square

Find an equation of the circle whose diameter has endpoints (-6,-3) and (-2,3) square
Find an equation of the circle whose diameter has endpoints (-6,-3) and (-2,3) square

Solution
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Answer

(x + 4)^2 + y^2 = 13 Explanation 1. Find the center of the circle The center (h, k) is the midpoint of the diameter endpoints (-6, -3) and (-2, 3) using h = \frac{x_1 + x_2}{2} and k = \frac{y_1 + y_2}{2}. h = \frac{-6 + (-2)}{2} = -4 k = \frac{-3 + 3}{2} = 0 Center: (-4, 0) 2. Calculate the radius squared The radius r is half the distance between the endpoints. Using the distance formula d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}, the diameter d is: d = \sqrt{(-2 - (-6))^2 + (3 - (-3))^2} = \sqrt{4^2 + 6^2} = \sqrt{16 + 36} = \sqrt{52} r = \frac{\sqrt{52}}{2} \implies r^2 = \frac{52}{4} = 13 3. Write the circle equation Substitute the center (h, k) = (-4, 0) and r^2 = 13 into the standard form **(x - h)^2 + (y - k)^2 = r^2**. (x - (-4))^2 + (y - 0)^2 = 13

Explanation

1. Find the center of the circle<br />The center $(h, k)$ is the midpoint of the diameter endpoints $(-6, -3)$ and $(-2, 3)$ using $h = \frac{x_1 + x_2}{2}$ and $k = \frac{y_1 + y_2}{2}$.<br />$h = \frac{-6 + (-2)}{2} = -4$<br />$k = \frac{-3 + 3}{2} = 0$<br />Center: $(-4, 0)$<br /><br />2. Calculate the radius squared<br />The radius $r$ is half the distance between the endpoints. Using the distance formula $d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$, the diameter $d$ is:<br />$d = \sqrt{(-2 - (-6))^2 + (3 - (-3))^2} = \sqrt{4^2 + 6^2} = \sqrt{16 + 36} = \sqrt{52}$<br />$r = \frac{\sqrt{52}}{2} \implies r^2 = \frac{52}{4} = 13$<br /><br />3. Write the circle equation<br />Substitute the center $(h, k) = (-4, 0)$ and $r^2 = 13$ into the standard form **$(x - h)^2 + (y - k)^2 = r^2$**.<br />$(x - (-4))^2 + (y - 0)^2 = 13$
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