How do you find the equation of the circle with a diameter that has endpoints at (-7, -2) and (-15, 6)?

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Answer 1 · 2015-12-28T15:19:01

${(x + 11)}^{2} + {(y - 2)}^{2} = 128$ $(x + 11)^2 + (y - 2)^2 = 128$

${(x - h)}^{2} + {(y - k)}^{2} = r^{2}$ $(x - h)^2 + (y - k)^2 = r^2$

Where $C : (h, k)$ $C: (h, k)$

To find the center, use the midpoint formula

$C : (\frac{x_{1} + x_{2}}{2}, \frac{y_{1} + y_{2}}{2})$ $C: ((x_1 + x_2)/2, (y_1 + y_2)/2)$

$\Rightarrow C : (\frac{- 7 + - 15}{2}, \frac{- 2 + 6}{2})$ $=> C: ((-7 + -15)/2, (-2 + 6)/2)$

$\Rightarrow C : (- 11, 2)$ $=> C: (-11, 2)$

To find the radius, use the distance formula

$r = \sqrt{{(x_{1} - x_{2})}_{1}^{2} + {(y_{1} - y_{2})}_{1}^{2}}$ $r = sqrt((x_1 - x_2)^2 + (y_1 - y_2)^2)$

$\Rightarrow r = \sqrt{{(- 7 - - 15)}^{2} + {(- 2 - 6)}^{2}}$ $=> r = sqrt((-7 - -15)^2 + (-2 - 6)^2)$

$\Rightarrow r = \sqrt{8^{2} + {(- 8)}^{2}}$ $=> r = sqrt(8^2 + (-8)^2)$

$\Rightarrow r = \sqrt{128}$ $=> r = sqrt(128$

$\Rightarrow r = 8 \sqrt{2}$ $=> r = 8sqrt2$

Therefore, the equation of the circle is

${(x + 11)}^{2} + {(y - 2)}^{2} = {(8 \sqrt{2})}^{2}$ $(x + 11)^2 + (y - 2)^2 = (8sqrt2)^2$

$\Rightarrow {(x + 11)}^{2} + {(y - 2)}^{2} = 128$ $=> (x + 11)^2 + (y - 2)^2 = 128$