Given Equations for Two Circles, How Does One Find the Coordinates of Their Point(s) of Intersection, If Any?

Given Equations for Two Circles, How Does One Find the Coordinates of Their Point(s) of Intersection, If Any?

Before attempting to create a method for solving this problem with any given circles, I’m going to attempt solving it for this arbitrary pair of circles. The small circle has radius 5, is centered at (-8,3), and has the equation (x+8)² + (y-3)² = 25. The larger circle has a radius of √85, is centered at (4,-3), and has the equation (x-4)² + (y+3)² = 85.

The first thing I will do is add 60 to each side of the equation of the smaller circle, changing that equation to (x+8)² + (y-3)² + 60 = 85. There are now two equations with 85 on one side, so I will set their other sides equal to each other:  (x+8)² + (y-3)² + 60 = (x-4)² + (y+3)². Next, I square each term in parentheses, turning this equation into x² + 16x +64 + y² -6y +9 + 60 = x² – 8x +16 + y² + 6y +9. After x², y², and 9 are each cancelled from each side, the equation looks like this:  16x +64 – 6y +60 = -8x + 16 + 6y. This simplifies further to -12y + 124= -24x +16, then -12y = -24x -108, then, after dividing both sides -12, that yields the simplified equation for the straight line y = 2x + 9. Any solutions which exist, therefore, must be found on the line y = 2x + 9.

(x+8)² + (y-3)² = 25 is the equation of the smaller circle, and substituting 2x + 9 for y in this equation turns it into (x+8)² + (2x + 9 – 3)² = 25, which is equivalent to (x+8)² + (2x + 6)² = 25. By squaring the terms in parentheses found in this last equation, this turns into x² + 16x +64 + 4x² + 24x + 36 = 25, which simplifies further to 5x² +40x + 75 = 0, and then x² +8x + 15 = 0.

Applying the quadratic formula (because I don’t enjoy factoring trimonials) to this last equation, x = {-8 ± sqrt[64 – (4)(1)15)]}/2, which means that x has two values in this case:  x = (-8 + 2)/2 = -6/2 = -3, and x = (-8 – 2)/2 = -10/2 = -5. When x = -3, then y = 2(-3) +9 = -6 + 9 = 3, so one solution is the ordered pair (-3,3). By constrast, when x = -5, then y = 2(-5) +9 = -10 + 9 = -1, so the other solution is the ordered pair (-5,-1).

For other circle-pairs, the method is the same, but appropriate substitutions will need to be made in the numbers with which I started, likely leading, of course, to different answers. For some pairs of circles, the quadratic formula will not yield two real ordered pairs as solutions. If all values of x and y are nonreal, this simply means that the two circles do not touch at all. If, on the other hand, only ordered pair can be found as a solution, this means the circles are tangent to each other — touching only at a single point.

About RobertLovesPi

I go by RobertLovesPi on-line, and am interested in many things. The majority of these things are geometrical. Welcome to my little slice of the Internet. The viewpoints and opinions expressed on this website are my own. They should not be confused with the views of my employer, nor any other organization, nor institution, of any kind.
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