Radical center and Monge's theorem have easy synthetic proofs. However, here we use coordinate approach to cover these cases:
- The intersections of two circles may be imaginary points, but there exists a straight line passing through them.
- The tangent lines to two circles may be imaginary lines, but their intersection point is real.
The radical line of two circles can be defined as the straight line passing through their intersections, whatever the two circles intersect or not. This is because any two circles can be denoted as:
Whatever the two roots are real or imaginary, we can always get a straight line by subtracting above two equations:
such that the two roots are on this line.
For example, one circle is inside the other:
The two intersections are imaginary: , but the straight line passing through them are real:
.
The radical lines of three circles are concurrent at a point called radical center or power center. [1]
Let's denote these three circles as:
Then the three radical lines are:
They are obviously concurrent because their determinant of coefficients is zero:
Here we are only interested in the intersection of two external tangent lines (later we call it external intersection) and the intersection of two internal ones (later we call it internal intersection), which means the intersection of an external one and an internal one is not considered.
Let's put two centers A and B onto y-axis, and denote these two circles as:
Because of the symmetry about y-axis, a pair of either external or internal tangent lines can be denoted as:
The intersection is (0, h), so we are only interested in h but not k.
When eliminating y with the first circle, we have:
which can be written as:
Because a tangent line has only one common point with each circle, the discriminant should be zero, i.e.
which can be written as:
Analogously, when tangent to the second circle, we have:
We get below equation by eliminating :
The first solution is:
which is less than b, and should be infinity if R = r (two tangent lines parallel to y-axis). So it should be the external intersection. Denote it as H, then we have:
The second solution is:
which is between a and b. So it should be the internal intersection. Denote it as H', then we have:
In fact, the external and internal intersections are the external and internal homothetic centers of two circles, and they are harmonic conjugate with respect to the two circle centers:
For example, one circle is inside the other:
The two external tangent lines are (tangent points are
and
), and their intersection (internal homothetic center) is
.
The two internal tangent lines are (tangent points are
and
), and their intersection (external homothetic center) is
.
Monge's theorem states that for any three circles in a plane, none of which is completely inside one of the others, the intersection points of each of the three pairs of external tangent lines are collinear.
In fact, there are another three propositions.
For any three circles (A), (B) and (C) in a plane, let D, E and F be the external homothetic centers (intersection points of external tangent lines, if one circle is not completely inside the other) of (B)(C), (C)(A) and (A)(B) respectively, and let D', E' and F' be the internal homothetic centers (intersection points of internal tangent lines, if the two circles are disjoint) of (B)(C), (C)(A) and (A)(B) respectively, then we have:
- DEF are collinear;
- DE'F' (D'EF'; D'E'F) are collinear;
- AD', BE' and CF' are concurrent;
- AD, BE and CF' (AD, BE' and CF; AD', BE and CF) are concurrent.
To prove these propositions, just apply Eq. 1 onto D, E and F, and apply Eq. 2 onto D', E' and F', then get collinearity by Menelaus's theorem, and get concurrency by Ceva's theorem.
A famous triangle center X(12) mentioned here (Theorem 3) can be proved by proposition 2.
- We use the diagram from here.