Radical axis

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Definitions

The power of point $A$ with respect to circle $\omega$ (with radius $r$ and center $O$ , which shall thereafter be dubbed $pow(P, \omega)$ , is defined to equal $OP^2 - r^2$ .

The radical axis of two circles $\omega_1, \omega_2$ is defined as the locus of the points $P$ such that the power of $P$ with respect to $\omega_1$ and $\omega_2$ are equal. In other words, if $O_i, r_i$ are the center and radius of $\omega_i$ , then a point $P$ is on the radical axis if and only if $PO_1^2 - r_1^2 = PO_2^2 - r_2^2$

Results

Theorem 1: (Power of a Point) If a line drawn through point P intersects circle $\omega$ at points A and B, then $pow(P, \omega) = PA * PB$ . Theorem 2: (Radical Axis Theorem)

a. The radical axis is a line perpendicular to the line connecting the circles' centers (line $l$ ).

b. If the two circles intersect at two common points, their radical axis is the line through these two points.

c. If they intersect at one point, their radical axis is the common internal tangent.

d. If the circles do not intersect, their radical axis is the perpendicular to $l$ through point A, the point on $l$ such that $pow(A, \omega_1) = pow(A, \omega_2)$ .

Theorem 3: (Radical Axis Concurrence Theorem) The three pairwise radical axes of three circles concur at a point, called the radical center.

Proofs

Theorem 1 is trivial Power of a Point, and thus is left to the reader as an exercise. (Hint: Draw a line through P and the center.)

Theorem 2 shall be proved here. Assume the circles are $\omega_1$ and $\omega_2$ with centers $O_1$ and $O_2$ and radii $r_1$ and $r_2$ , respectively. (It may be a good idea for you to draw some circles here.)

Lemma 1: Let $P$ be a point in the plane, and let $P'$ be the foot of the perpendicular from $P$ to $O_1O_2$ . Then $pow(P, \omega_1) - pow(P, \omega_2) = pow(P', \omega_1) - pow(P', \omega_2)$ .

The proof of the lemma is an easy application of the Pythagorean Theorem and will again be left to the reader as an exercise.

Lemma 2: There is an unique point P on line $O_1O_2$ such that $pow(P, O_1) = pow(P, O_2)$ .

Proof: First show that P lies between $O_1$ and $O_2$ . Then, use the fact that $O_1P + PO_2 = O_1O_2$ (a constant) to prove the lemma.

Lemma 1 shows that every point on the plane can be equivalently mapped to a line on $O_1O_2$ . Lemma 2 shows that only one point in this mapping satisfies the given condition. Combining these two lemmas shows that the radical axis is a line, completing part (a).

Parts (b), (c), and (d) will be left to the reader as an exercise.

Now, try to prove Theorem 3 on your own!

Exercises

If you haven't already done so, prove the theorems and lemmas outlined in the proofs section. If you have done so, prove them again. (Repetition is the best way to master a new concept.)

Note: No solutions will be provided to the following problems. If you are stuck, ask on the forum.

Problem 1. Two circles P and Q intersect at X and Y. Point A is located on $XY$ such that AP = 10 and AQ = 15. If the radius of Q is 7, find the radius of P.

Problem 2. Solve 2009 USAMO Problem 1. If you already know how to solve it, write up a real proof on a piece of paper. And do it twice.

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Radical axis

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