Find the equation of the circle such that the three points A(- 5 , 0), B(1 , 0) and D(- 2 , - 3) are on the circle.

Solution

Let (h , k) be the center and r be the radius of the circle and write the general equation of the circle as follows
(x - h)^{2} + (y - k)^{2} = r^{2} The three points are on the circle and therefore satisfy the above equation of the circle. Hence the three equations obtained by substituting the values of the coordinates x and y of the points into the equation
(-5 - h)^{2} + (0 - k)^{2} = r^{2}
(1 - h)^{2} + (0 - k)^{2} = r^{2}
(- 2 - h)^{2} + (- 3 - k)^{2} = r^{2} We now need to solve the above non linear system of equations in three variable. Expand the squares in the above equations and rewrite them as follows
25 + 10 h + h^{2} + k^{2} = r^{2} (1)
1 - 2 h + h^{2} + k^{2} = r^{2} (2)
4 + 4h + h^{2} + 9 + 6 k + k^{2} = r^{2} (3)
Subtract the left and right hand sides of equations (1) and (2) to obtain another equivalent equation
(25 + 10 h + h^{2} + k^{2}) - (1 - 2 h + h^{2} + k^{2}) = r^{2} - r^{2} Simplify to obtain a simpler equation
25 + 10 h - 1 + 2h = 0
Solve the above linear equation to obtain
h = - 2
Subtract the left and right hand sides of equations (2) and (3) to obtain another equivalent equation
(1 - 2 h + h^{2} + k^{2}) - (4 + 4h + h^{2} + 9 + 6 k + k^{2}) = r^{2} - r^{2} Simplify to obtain the linear equation
- 6 h - 6 k - 12 = 0
Substitute h by -2 in the above equation and solve for k
- 6 (-2) - 6 k - 12 = 0
k = 0
Substitute h by - 2 and k by 4 in any of the three equations (1), (2) or (3) and solve for r. Let us use equation (1)
25 + 10 (-2) + (-2)^{2} + (0)^{2} = r^{2}
r^{2} = 9
r = 3
Substitute h, k and r by their values found above into the general equation to obtain the equation of the circle through the three given points.
(x + 2)^{2} + y^{2} = 3^{2}

In the figure below are shown the given circle and the given three points and we can see that the points are on the circle.