How do you find the equation of a circle with diameter has endpoints (-2, 3) and (4, -1)?

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Answer 1 · 2016-08-14T20:39:17

#(x-1)^2+(y-1)^2 = 13#

The radius of the circle is half of the length of the diameter, so half the distance between these two endpoints:

#1/2 sqrt((4-(-2))^2+((-1)-3)^2) = 1/2 sqrt(36+16) = 1/2 sqrt(52) = sqrt(13)#

The centre of the circle is the midpoint of the diameter:

#(((-2)+4)/2, (3+(-1))/2) = (1, 1)#

The equation of a circle with centre #(h, k)# and radius #r# may be written:

#(x-h)^2+(y-k)^2 = r^2#

So in our case:

#(x-1)^2+(y-1)^2 = 13#

graph{((x-1)^2+(y-1)^2-13)((x-1)^2+(y-1)^2-0.02)((x+2)^2+(y-3)^2-0.02)((x-4)^2+(y+1)^2-0.02) = 0 [-10, 10, -5, 5]}