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

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Answer 1 · 2017-08-02T12:28:00

The equation of the circle is ${(x + 11)}^{2} + {(y - 2)}^{2} = 32$ $(x+11)^2+(y-2)^2=32$

The points on the diameter are $A = (- 7, - 2)$ $A=(-7,-2)$ and $B = (- 15, 6)$ $B=(-15,6)$

The midpoint of $A B$ $AB$ is the center of the circle

$O = (\frac{(- 7) + (- 15)}{2}, \frac{(- 2) + (6)}{2}) = (- 11, 2)$ $O=(((-7)+(-15))/2, ((-2)+(6))/2)=(-11,2)$

The radius of the circle is

$r^{2} = ({(- 7 + 11)}^{2} + {(- 2 - 2)}^{2}) = (4^{2} + 4^{2}) = 32$ $r^2=((-7+11)^2+(-2-2)^2)=(4^2+4^2)=32$

The equation of the circle is

${(x + 11)}^{2} + {(y - 2)}^{2} = r^{2} = 32$ $(x+11)^2+(y-2)^2=r^2=32$

graph{(x+11)^2+(y-2)^2=32 [-23.04, 2.27, -3.43, 9.24]}