Theorem 3.13: Given two circles whose equations are
their points of intersection are given by
Proof: The points of intersection lie on a line whose equation is given from Theorem 3.10.
We will consider two cases. The first is where the centers are not horizontal, and the second is where they are.
Case 1: The centers are not horizontal. We can use the simpler formula for the points where a line intersects a circle from Theorem 3.3
where by Theorem 3.10,
The rational terms are the coordinates of the foot of the center of the circle centered at (x1, y1) with radius r1. The slope of the line between the points of intersection of the two circles is the negative reciprocal of the slope of the line between the centers we know that the line joining the points of intersection is perpendicular to the line between the centers, so the foot of the center in the line joining the two points is the intesection of those lines, and we know from Theorem 3.12 that those coordinates are
so the coordinates of the points of intersection are
We should now work with the radical terms. Let us concentrate on the expression under the radicals
First note that
When we substitiute for m and b, the expression inside the radical becomes
Find common denominators.
Remove the parentheses in the second term.
Combine like terms in the numerator of the second fraction.
This can be rewritten as
We we find common denominators for these two fractions.
We can now factor the top as a difference of squares.
Remove the inside parentheses, and rearrange the terms.
Both factors on top are differences of squares, and will factor.
Which can be written as
Rearrange the factors
and we find that we have two sum times difference multiplications which we can multiply as
We put this back under the radical in the formula for x and using the fact that
for the radical term. Invert and multiply the denominator, and simplify the denominator under the radical.
If we put it all together, we get
Since we will be taking both the positive and negative square root, we can drop the absolute value signs around y2 - y1.
To get the y-coordinate of the points of intersection we can substitute this value of x into the equation from Theorem 3.10. This would give us
when you multiply m times x, you will need to multiply m by all the terms, but when you add b, you will only be adding it in once. When you multiply m times the rational terms and add b, you will get the y-coordinate of the foot of the (x1, y1) in the line joining the points of intersection of the circles, but after that, you will simply multiply the radical term by m. Since
the y2 - y1's will cancel leaving us with an x2 - x1 on top. The negative sign in front of the slope will change th sign resulting in
Case 2: y1 = y2. In this case the line which contains the points of intersection has the equation
by Theorem 3.12. Subtitute this value for x into the equation for the first circle
(x - x1)2 + (y - y1)2 = r12
This will simplify to
This is fairly easy to solve for y. Transpose the first term on the left to the right.
Take square roots of both sides.
To simplify this, first find common denominators for the terms inside the parentheses under the radical.
Under the condition that y1 = y2, |x2 - x1| = d, the distance between the centers of the circles.After finding common denominators for the two terms inside the radical, we get
The expression under the radical is essentially the same as the expression under the radical in the last computation, so it will simplify to the same thing.
which is the formula to which the y coordinates of the points of intersection of the circles, in the statement of the theorem, will simplify, when y1 = y2.
Next theorem (T3.14)