How do you find the equation of a circle that passes through the following points: (5,3) (-2,2) (-1,-5)?

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Answer 1 · 2016-07-28T21:41:42

#(x-2)^2+(y+1)^2 = 5^2#

Notice that:

#(-2, 2) - (5, 3) = (-7, -1)#

#(-1, -5) - (-2, 2) = (1, -7)#

So the line segment joining #(5, 3)# and #(-2, 2)# is orthogonal to and the same length as the line segment joining #(-2, 2)# and #(-1, -5)#.

Thus these three points form an isosceles triangle with angles #pi/4#, #pi/4# and #pi/2#.

The centre of the circle passing through them is therefore the midpoint of the line segment joining #(5, 3)# and #(-1, -5)#, namely:

#((5-1)/2, (3-5)/2) = (2, -1)#

The radius is the distance from this point to any of the original three points, say #(5, 3)#:

#r = sqrt((5-2)^2+(3-(-1))^2) = sqrt(3^2+4^2) = sqrt(9+16) = sqrt(25) = 5#

The equation of the circle can therefore be written:

#(x-2)^2+(y+1)^2 = 5^2#

graph{((x-2)^2+(y+1)^2-25)((x-2)^2+(y+1)^2-0.02)((x+2)^2+(y-2)^2-0.04)((x-5)^2+(y-3)^2-0.04)((x+1)^2+(y+5)^2-0.04) = 0 [-9.29, 10.71, -5.96, 4.04]}