Math Core

Whenever a line through a point $P$ meets a circle twice, the product of the two distances from $P$ doesn't depend on which line you chose. That single fact, the power of a point, turns chords, secants and tangents into algebra, and it leads straight to the radical axis of two circles, one of the slickest tools on the AIME.

The power of a point

Definition

Power of a point

For a circle with center $O$ and radius $r$ , the power of a point $P$ is

$\operatorname{pow}(P) = OP^2 - r^2.$

It is positive outside the circle, zero on it, and negative inside.

Power of a point theorem

If a line through $P$ meets the circle at $A$ and $B$ , then

$PA \cdot PB = |\operatorname{pow}(P)| = |OP^2 - r^2|.$

If $P$ is outside and $PT$ is tangent at $T$ , then $PT^2 = OP^2 - r^2$ too.

This packs three familiar rules into one:

Two chords $AB$ and $CD$ crossing at $P$ inside: $PA \cdot PB = PC \cdot PD$ .
Two secants from $P$ outside: $PA \cdot PB = PC \cdot PD$ (whole times outside part).
Tangent and secant from $P$ : $PT^2 = PA \cdot PB$ .

Proof. Put $O$ at the origin and write the line as $P + t\,\mathbf{u}$ with $\mathbf{u}$ a unit vector. The point is on the circle when $|P + t\mathbf{u}|^2 = r^2$ , that is,

$t^2 + 2(P \cdot \mathbf{u})\,t + \left(|P|^2 - r^2\right) = 0.$

The two roots $t_1, t_2$ are the signed distances from $P$ to $A$ and $B$ , and by Vieta their product is $|P|^2 - r^2 = OP^2 - r^2$ , no matter which direction $\mathbf{u}$ you picked. A tangent line gives a double root $t_1 = t_2 = PT$ , so $PT^2 = OP^2 - r^2$ . $\blacksquare$

(There's also a classic similar-triangles proof: for crossing chords, $\triangle PAC \sim \triangle PDB$ because inscribed angles on the same arc are equal, so $\dfrac{PA}{PD} = \dfrac{PC}{PB}$ .)

Two chords AB and CD crossing at P inside the circle: PA · PB = PC · PD.

Worked example: Crossing chords

Chords $AB$ and $CD$ meet at $P$ , with $PA = 4$ , $PB = 9$ and $PC = PD$ . Find $CD$ .

$PC \cdot PD = PA \cdot PB = 36$ , so $PC = PD = 6$ and $CD = 12$ .

Worked example: Tangent and secant

From $P$ , a tangent touches a circle at $T$ with $PT = 12$ , and a secant meets the circle at $A$ and $B$ with $PA = 8$ . Find $AB$ .

$PT^2 = PA \cdot PB$ gives $144 = 8 \cdot PB$ , so $PB = 18$ and $AB = 18 - 8 = 10$ .

The radical axis

Two circles give every point two powers. Where are they equal?

Radical axis

For two circles with different centers, the set of points with equal power to both is a line, the radical axis. It is perpendicular to the line of centers. If the circles meet, it is the line through their two intersection points (the common chord, extended).

Proof. Write the circles as $x^2 + y^2 + D_1x + E_1y + F_1 = 0$ and $x^2 + y^2 + D_2x + E_2y + F_2 = 0$ . Expanding $OP^2 - r^2$ shows that the power of $(x, y)$ with respect to each circle is just the left side of its equation. Setting the powers equal, the $x^2 + y^2$ cancels:

$(D_1 - D_2)x + (E_1 - E_2)y + (F_1 - F_2) = 0,$

a line. Its normal vector $(D_1 - D_2, E_1 - E_2)$ is a multiple of the vector between the centers $\left(-\tfrac{D_1}{2}, -\tfrac{E_1}{2}\right)$ and $\left(-\tfrac{D_2}{2}, -\tfrac{E_2}{2}\right)$ , so the line is perpendicular to the line of centers. An intersection point has power $0$ to both circles, so it's on the line. $\blacksquare$

Two very useful consequences:

Equal tangents. From any point on the radical axis (outside the circles), the tangent segments to the two circles are equal.
Radical center. For three circles with non-collinear centers, the three radical axes meet at one point. (A point with equal power to circles $1$ and $2$ and to circles $2$ and $3$ also has equal power to $1$ and $3$ .) So three pairwise common chords are always concurrent.

Two circles and their radical axis x = 4, which passes through both intersection points.Open in grapher →

Worked example: Common chord by subtraction

Find the length of the common chord of $x^2 + y^2 = 25$ and $(x - 6)^2 + y^2 = 13$ .

Subtract the equations: $x^2 - (x - 6)^2 = 12$ , so $12x - 36 = 12$ and $x = 4$ . That's the radical axis. On it, $y^2 = 25 - 16 = 9$ , so the circles meet at $(4, \pm 3)$ and the common chord has length $6$ .

Worked example: The radical axis bisects a common tangent

Two circles meet at $P$ and $Q$ . A common tangent touches them at $A$ and $B$ . Show that line $PQ$ passes through the midpoint of $AB$ .

Let line $PQ$ meet $AB$ at $M$ . Since $M$ is on the radical axis, its powers are equal: $MA^2 = MP \cdot MQ = MB^2$ . So $MA = MB$ .

Common mistake

Power of a point multiplies distances from $P$ , not pieces of the chord. For a secant from outside, it's $PA \cdot PB$ where $PB$ is the whole length from $P$ to the far point, not the chord $AB$ .

Tip

When an AIME problem has two or three circles and asks about a line through their intersections, think radical axis first. And when a circle passes through a vertex and is tangent to a side, power of a point from the other vertices usually gives lengths immediately.

Practice

Practice 1

Chords $AB$ and $CD$ of a circle meet at $P$ . Given $AP = 6$ , $PB = 10$ and $CD = 19$ , find $CP^2 + PD^2$ .