**Theorem
3.12**: The point
where the line joining the
centers of two circles meets the
line containing the
points of intersection of
the two circles is given by

and

where (*x*_{1}, *y*_{1}) and (*x*_{2},
*y*_{2}) are the centers of the circles and *r*_{1 }and *r*_{2 }are the respective
radii, and *d *is the distance between the centers.

**Proof**: **Case 1**: *x*_{1} = *x*_{2}. In
this case the equation of the line joining the centers is *x* =
*x*_{1}, and *d* = |*y*_{2} - *y*_{1}|. From Theorem 3.10, the equation
containing the points of
intersection will be

after we simplify the formula by taking into account that *x*_{1} = *x*_{2}. In this case the point of intersection will
be

to which, in this case, the formula, in the statement of the theorem, will simplify.

**Case 2**: *y*_{1} = *y*_{2}. In this case the
equation of the line joining the centers is *y* = *y*_{1}, and *d* = |*x*_{2} -* x*_{1}|. From Theorem 3.10, the equation containing the
points of intersection
will be

so the point of intersection will be

If *y*_{1} = *y*_{2}, then the formulas in the
statement of the theorem simplify to give these coordinates.

**Case 3**: The line
joining the centers is neither
horizontal nor
vertical. Since the
line containing the
solutions is
perpendicular to
the line joining the
centers, by Theorem 3.11, we know the
point of intersection will
be the foot of (*x*_{1}, *y*_{1}) in the line containing the
solutions, which by Theorem 3.10 is

We can then use Theorem 1.10 to get the coordinates of the foot as

where

and

Substituting this in to the formula for the *x*-coordinate of the foot gives us

First let find common denominators for the first two terms on the top, clean up the last term, and find commondenominators for the bottom.

Remove parentheses on the top of the first term in the numerator.

After we cancel the like terms, let us rearrange the terms in the second factor of the top of the second fraction on the top.

To clear denominators in the compound fraction, we multiply the
top and bottom by 2(*y*_{2} - *y*_{1})^{2}.

In the second term on top, let us factor the difference of the squares in the *y*_{2}^{2} - *y*_{1}^{2} and *x*_{2}^{2} - *x*_{1}^{2}.

We can now factor out *y*_{2} - *y*_{1 }in the first
two terms in the top and clean up the third term.

Remove parentheses inside the factor that comes after the *y*_{2} - *y*_{1}.

Combine like terms

The second factor in the first term factors

Factor out the *x*_{2} + *x*_{1} from the first two terms.

We can split this up into

This simplifies to

We could use a similar approach to find the *y*-coordinate, but it
might be simpler to use the fact that the point is on the
line joining the centers,
and using the equation for the
line that joins (*x*_{1}, *y*_{1}) and (*x*_{2}, *y*_{2}), which, since we have voided the case where the centers of the circle are
vertical, is of the
form *y* = *mx* + *b *by Theorem 1.2.

This becomes

or

In the first two terms, we are substituting the *x*-coordinate for
the midpoint of the
line segment
between (*x*_{1}, *y*_{1}) and (*x*_{2},
*y*_{2}) into the equation for the line that they determine,
and so we should get the *y-*coordinate of that same midpoint which by
Theorem 2.4 is

By Theorem 1.4

This gives us

or