**Theorem
3.13**: Given two circles whose
equations are

and

their points of intersection are given by

and

where

**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

and

where by Theorem 3.10,

and

The rational terms are the coordinates of the foot of the center of the circle centered at (*x*_{1}, *y*_{1}) with radius *r*_{1}. 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

and

so the coordinates of the points of intersection are

and

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.

or

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

get

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 *y*_{2} - *y*_{1}.

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 (*x*_{1}, *y*_{1}) 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 *y*_{2} - *y*_{1}'s will cancel leaving us with an *x*_{2} - *x*_{1 }on top. The negative sign in front of the slope will change th sign resulting in

**Case 2**: *y*_{1} = *y*_{2}. 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* - *x*_{1})^{2} + (*y* - *y*_{1})^{2} = *r*_{1}^{2}

and get

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.

or

To simplify this, first find common denominators for the terms inside the parentheses under the radical.

Under the condition that *y*_{1} = *y*_{2}, |*x*_{2} - *x*_{1}| = *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 *y*_{1} = *y*_{2}.