The Radical Axis

Radical Axis

Once we have the power of a point with respect to a circle, the natural question to ask is what happens if we have two circles? What is the relation of power of a point with respect to one circle and power of the same point with respect to another circle?

Particularly, we are interested in finding out when the two powers are equal, or the locus of $P$ so that power of $P$ with respect to two circles $S_1$ and $S_2$ is equal. This also means that the tangents to both circles from point $P$ will have equal length.

Once you construct the above locus for same tangent length and for the circles $S_1$ and $S_2$, you can take their intersection and get a few points which have equal power with respect to both the circles!

Lets try to find the locus for two non-concentric circles.

Now that we have a claim, lets try to prove it!

Here, $\overline{PX} \perp \overline{O_1O_2}$

So we have a strategy. And we also have the power of point expression: $O_1P^2 - r_1^2 = O_2P^2 - r_2^2$.

Pythagoras theorem gives us that $O_1P^2 = O_1X^2 + PX^2$ and $O_2P^2 = O_2X^2 + PX^2$ . Substituting back in the equation, we get that $O_1X^2 - O_2X^2 = r_1^2 - r_2^2$

As we were dealing with squares till now, we can take all distances to be directed and show that such point $X$ is unique on line $\overline{O_1O_2}$.

And hence we have that $X$ is independent of $P$. Which tells us that $P$ always lies on a line perpendicular to $\overline{O_1O_2}$ passing through $X$.

Hence, the locus of $P$ must be a line.

We can retrace our steps in above proof to complete proof of part 2. The Pythagoras theorem finishes the problem.

We define this particular line to be the Radical Axis of two circles. Note that this definition is only valid when two circles and non-concentric.

Coordinate Geometry Proof

This proof uses coordinate geometry. It is a very elegant proof

Now we just have to equate two such equations!!

Every point $P(x_0 , y_0)$ having equal powers with respect to both the circles satisfies $s_1(x_0, y_0) - s_2(x_0 , y_0)$ and every point satisfying this equation will have equal power with respect to both the circles! Hence, we directly get the locus given by $s_1(x,y) - s_2(x,y) = 0$ which is generally of the form $ax + by + c =0$, which represents a line. The special case when both $a = 0$ and $b = 0$ represent concentric circles.

Remark : Here both the equations must be of type $x^2 + y^2 + 2gx + 2fy + c = 0$

Hence we are done! You can also do a simple computation to prove that this line is perpendicular to line joining the centers.

Properties of Radical Axis

As we know, Radical Axis is perpendicular to line joining the centers

Expressing power of point for tangent case gives us that tangents from any point on radical axis to both the circles have equal lengths.

Thus, Radical Axis is also the locus of center of a circle which is orthogonal to both the given circles

Degenerate Case

Radical axis of two point circles is the perpendicular bisector. In fact, we can take two congruent circles and their radical axis will be the perpendicular bisector of the centers.