How do you write an equation for a circle given endpoints of a diameter at (-5,2) and (3,6)?

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Answer 1 · 2016-10-24T04:58:50

${(x + 1)}^{2} + {(y - 4)}^{2} = 20$ $(x+1)^2+(y-4)^2=20$

First, we have to find the centre of the circle.

This will be halfway between the two endpoints

$\frac{3 + (- 5)}{2} = - 1$ $(3+(-5))/2 =-1$ , $\frac{6 + 2}{2} = 4$ $(6+2)/2=4$

giving us the point $(- 1, 4)$ $(-1,4)$ for the centre

now we need to find the radius,
take the middle and an end point and find the distance,

$((3) - (- 1)$ $((3)-(-1)$ , $(6) - (4))$ $(6)-(4))$

giving us $(4, 2)$ $(4,2)$ or 4 our units horizontal and two vertical.

now if we use trigonometry to find the radius

$r = \sqrt{4^{2} + 2^{2}} = \sqrt{20}$ $r=sqrt(4^2+2^2) = sqrt(20)$

now the equation of a circle is,

${(x - a)}^{2} + {(y - b)}^{2} = r^{2}$ $(x-a)^2+(y-b)^2=r^2$

where a and b are the coordinates of the centre.
sub in,

${(x - (- 1))}^{2} + {(y - (4))}^{2} = {(\sqrt{20})}^{2}$ $(x-(-1))^2+(y-(4))^2=(sqrt(20))^2$

${(x + 1)}^{2} + {(y - 4)}^{2} = 20$ $(x+1)^2+(y-4)^2=20$

giving us,

${(x + 1)}^{2} + {(y - 4)}^{2} = 20$ $(x+1)^2+(y-4)^2=20$

for the equation of our circle.