This is tutorial on finding the points of intersection of two circles given by their equations; general solution.
Example 1: Find the points of intersection of the circles given by their equations as follows:
(x  2)^{2} + (y  3)^{2} = 9
(x  1)^{2} + (y + 1)^{2} = 16
Solution to Example 1:
 We first expand the two equations as follows:
x^{2}  4x + 4 + y^{2}  6y + 9 = 9
x^{2}  2x + 1 + y^{2} + 2y + 1 = 16
 Multiply all terms in the first equation by 1 to abtain an equivalent equation and keep the second equation unchanged
x^{2} + 4x  4  y^{2} + 6y  9 = 9
x^{2}  2x + 1 + y^{2} + 2y + 1 = 16
 We now add the same sides of the two equations to obtain a linear equation
2x  3 + 8y  8 = 7
 Which may written as
x + 4y = 9 or x = 9  4y
 We now substitute x by 9  4y
in the first equation to obatin
(9  4y)^{2}  4(9  4y) + 4 + y^{2}  6y + 9 = 9
 Which may be written as
17y^{2} 62y + 49 = 0
 Solve the quadratic equation for y to obtain two solutions
y = (31 + 8√2) / 17 ≈ 2.49
and y = (31  8√2) / 17 ≈ 1.16
 We now substitute the values of y already obtained into the equation x = 9  4y to obtain the values for x as follows
x = (29 + 32√2) / 17 ≈  0.96
and x = (29  32√2) / 17 ≈ 4.37
 The two points of intersection of the two cirlces are given by
( 0.96 , 2.49) and (4.37 , 1.16)
Shown below is the graph of the two circles and the linear equation x + 4y = 9.
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