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What is the standard form of the equation of a circle with endpoints of a diameter at the points (7,8) and (-5,6)?

1 Answer

Ananda Dasgupta

Mar 6, 2018

${(x - 1)}^{2} + {(y - 7)}^{2} = 37$ $(x-1)^2+(y-7)^2=37$

Explanation:

The center of the circle is the midpoint of the diameter, i.e. $(\frac{7 - 5}{2}, \frac{8 + 6}{2}) = (1, 7)$ $((7-5)/2, (8+6)/2) = (1,7)$

Again, the diameter is the distance between the points s $(7, 8)$ $(7,8)$ and $(- 5, 6)$ $(-5,6)$ :

$\sqrt{{(7 - (- 5))}^{2} + {(8 - 6)}^{2}} = \sqrt{12^{2} + 2^{2}} = 2 \sqrt{37}$ $sqrt((7-(-5))^2+(8-6)^2) = sqrt(12^2+2^2)=2sqrt(37)$

so the radius is $\sqrt{37}$ $sqrt(37)$ .

Thus the standard form of the circles equation is

${(x - 1)}^{2} + {(y - 7)}^{2} = 37$ $(x-1)^2+(y-7)^2=37$

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