How do you write an equation of a circle whose diameter has endpoints (-2, 3) and (4, -1)?

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How do you write an equation of a circle whose diameter has endpoints (-2, 3) and (4, -1)?

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Answer 1 · 2016-08-23T16:33:55

$x^{2} + y^{2} - 2 x - 2 y - 11 = 0$ $x^2+y^2-2x-2y-11=0$ in the standard form

Explanation:

To obtain the equation of the circle, you need to know its center and radius; the center is the middle point of the given points and the radius is the half of the diameter, so the center is

$M (\frac{- 2 + 4}{2}; \frac{3 - 1}{2}) = (1; 1)$ $M((-2+4)/2;(3-1)/2)=(1;1)$

and the radius is

$r = \frac{1}{2} \sqrt{{(4 + 2)}^{2} + {(- 1 - 3)}^{2}} = \frac{1}{2} \sqrt{36 + 16} = \frac{1}{2} \sqrt{52} = \sqrt{13}$ $r=1/2sqrt((4+2)^2+(-1-3)^2)=1/2sqrt(36+16)=1/2sqrt(52)=sqrt(13)$

Then let's substitute them in the equation:

${(x - x_{0})}_{0}^{2} + {(y - y_{0})}_{0}^{2} = r^{2}$ $(x-x_0)^2+(y-y_0)^2=r^2$

to have

${(x - 1)}^{2} + {(y - 1)}^{2} = 13$ $(x-1)^2+(y-1)^2=13$

$x^{2} + y^{2} - 2 x - 2 y + 1 + 1 - 13 = 0$ $x^2+y^2-2x-2y+1+1-13=0$

$x^{2} + y^{2} - 2 x - 2 y - 11 = 0$ $x^2+y^2-2x-2y-11=0$ in the standard form

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How do you write an equation of a circle whose diameter has endpoints (-2, 3) and (4, -1)?

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