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   (x1, y1)   and   (x2, y2)   are the centers of the circles and   r1   and   r2   are the respective radii, and   d   is the distance between the centers.

Proof: Case 1:   x1 = x2.   In this case the equation of the line joining the centers is   x = x1,   and   d = |y2 - y1|.   From Theorem 3.10, the equation containing the points of intersection will be

after we simplify the formula by taking into account that   x1 = x2.   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:   y1 = y2.   In this case the equation of the line joining the centers is   y = y1,   and   d = |x2 - x1|.   From Theorem 3.10, the equation containing the points of intersection will be

so the point of intersection will be

If   y1 = y2,   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   (x1, y1)   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(y2 - y1)2.

In the second term on top, let us factor the difference of the squares in the   y22 - y12   and   x22 - x12.

We can now factor out   y2 - y1   in the first two terms in the top and clean up the third term.

Remove parentheses inside the factor that comes after the   y2 - y1.

Combine like terms

The second factor in the first term factors

Factor out the   x2 + x1   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   (x1, y1)   and   (x2, y2),   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   (x1, y1)   and   (x2, y2)   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

This gives us

or

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next theorem (3.13)