How do you write an equation for a circle with endpoints of the diameter at (2, -5) and (-4, 3)?

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Answer 1 · 2016-08-11T07:03:02

$x^2+y^2+2x+2y-23=0$

To obtain the radius you first can find the lenght of the diameter by calculating the distance between its endpoints:

$d=sqrt((x_2-x_1)^2+(y_2-y_1)^2)$

$d=sqrt((3+5)^2+(-4-2)^2)$

$=sqrt(64+36)=10$

the radius is then the half of the diameter:

$r=5$

The center of the circle is the middlepoint of the diameter:

$M=((x_1+x_2)/2;(y_1+y_2)/2)$

$M=((2-4)/2;(-5+3)/2)=(-1;-1)$

The equation of the circle is obtained by substituting the center and the radius in the following:

$(x-x_0)^2+(y-y_0)^2=r^2$

where the center is $(x_0,y_0)=(-1)-1)$

and the radius is 5:

$(x+1)^2+(y+1)^2=5^2$

$x^2+y^2+2x+2y+1+1-25=0$

in the standard form:

$x^2+y^2+2x+2y-23=0$