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

##### 1 Answer

#### Explanation:

Notice that:

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

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

So the line segment joining

Thus these three points form an isosceles triangle with angles

The centre of the circle passing through them is therefore the midpoint of the line segment joining

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

The radius is the distance from this point to any of the original three points, say

#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]}