Overview

Power of a point is a key strategy when working with circles, lines, and points that shows up often in AMC/AIME problems. It's often applicable when multiple lines sharing one point intersect a circle, such as secants, tangents, or intersecting chords. The shared point may be inside or outside the circle, each yielding different formulas. The idea is that certain products of segment lengths are equal, allowing you to turn a geometric setup into a simple algebra equation and quickly find missing lengths.

Definitions

• A chord is a line segment with both endpoints on the circle.
• A secant line is a line that intersects a circle at two points.
• A tangent line is a line that touches a circle at exactly one point. When a radius is drawn from the center to this point, it will be perpendicular to the original tangent line.

Cases

Power of a point changes slightly depending on whether the point of intersection is inside or outside the circle, and whether the lines through the point are tangent or secant lines.

Case 1: Inside (two intersecting chords)

If chords $AB$ and $CD$ intersect at point $P$ inside a circle, then it holds that

Proof: Observe that $\angle ACD = \angle DBA$ as they are inscribed in the same arc. Similarly, $\angle CAB = \angle BDC,$ so $\triangle ACP \sim \triangle DBP.$ Then, we obtain

Case 2: Outside

If line $PX$ is tangent to the circle, and $PB$ and $PC$ are secants intersecting the circle at $A$ and $D,$ respectively, then it holds that

Note: $PX$ can be thought of as just another secant, except the two points at which it intersects the circle have become so close that they have condensed into one point, $X.$ From here the product $(PA)(PB)$ or similar becomes $(PX)(PX),$ revealing why the $PX^2$ term appears.

Proof: We will prove the case with two secants. Observe that $\angle ABC + \angle ADC = 180^\circ,$ as the arcs they are inscribed in form a $360^\circ$ circle. Therefore, $\angle ADP = 180 - \angle ADC = \angle ABC.$ Finally, it is clear that $\angle APD = \angle BPC,$ so we have $\triangle APD \sim \triangle CPB.$ Therefore,

The proof of the case involving tangent lines is left as an exercise.

Applications

Identify the Configuration

If you see two intersecting secants, chords, or tangents, consider applying power of a point.

Combine with Triangle Similarity or Angle Chasing

Triangle similarity and angle chasing were the two key strategies used in the proof of power of a point.

Alternate Form

If the distance from point $P$ to center $O,$ and radius $r$ are known, we can construct a secant/chord. Then, the power of point $P$ equals $|OP^2-r^2|,$ the absolute value included due to the two cases of $P$ being inside or outside the circle. Note that this expression is equal to $|(OP-r)(OP+r)|,$ making the geometric significance clearer.

In the below diagram, $(PA)(PB) = (r-OP)(r+OP) = r^2-OP^2 = (PC)(PD).$

In the below diagram, $(PA)(PB) = (OP-r)(OP+r) = OP^2-r^2 = (PC)(PD).$

Common Mistakes

• Mixing up order along a secant: In the case where $P$ is outside the circle, ensure you are multiplying $PA \cdot PB,$ and not $PA \cdot AB$ or similar. Be especially cautious of this mistake when the lengths for $PA$ and $AB$ are marked, as it may be tempting to multiply by $AB$ instead of $PB.$

• Mixing up secant and tangent: Accidentally squaring a secant length (e.g. $PA^2$ instead of using the product $(PA)(PB).$)
• Incorrectly applying: Applying power of a point when two segments through $P$ appear to be part of the same line, but are actually not.

Worked Example

A circle has radius $5$. Point $P$ is $8$ units from the center. If a secant through $P$ meets the circle at $A$ and $B$ with $PA=2$, find $PB$.

Power is $8^2-5^2=39$. So $2\cdot PB = 39$, giving $PB=39/2$.

More Examples

Example 1: Tangent Length

Point $P$ is $13$ units from the center of a circle of radius $5$. Find the tangent length.

Solution: $PT^2=13^2-5^2=144$, so $PT=12$.

Example 2: Two Secants

From $P$, one secant has $PA=3$ and $PB=15$. Another secant has $PC=5$. Find $PD$.

Solution: $3\cdot 15 = 5\cdot PD$, so $PD=9$.

Example 3: 2020 AMC 12B Problems/Problem 12

Let $\overline{AB}$ be a diameter in a circle of radius $5\sqrt2.$ Let $\overline{CD}$ be a chord in the circle that intersects $\overline{AB}$ at a point $E$ such that $BE=2\sqrt5$ and $\angle AEC = 45^{\circ}.$ What is $CE^2+DE^2? \\$

$\textbf{(A)}\ 96 \qquad\textbf{(B)}\ 98 \qquad\textbf{(C)}\ 44\sqrt5 \qquad\textbf{(D)}\ 70\sqrt2 \qquad\textbf{(E)}\ 100$

Let $O$ be the center of the circle, and $X$ be the midpoint of $CD$. Draw triangle $OCD$, and median $OX$. Because $OC = OD$, $OCD$ is isosceles, so $OX$ is also an altitude of $OCD$. $OE = 5\sqrt2 - 2\sqrt5$, and because angle $OEC$ is $45$ degrees and triangle $OXE$ is right, $OX = EX = \frac{5\sqrt2 - 2\sqrt5}{\sqrt2} = 5 - \sqrt{10}$. Because triangle $OXC$ is right, $CX = \sqrt{(5\sqrt2)^2 - (5 - \sqrt{10})^2} = \sqrt{15 + 10\sqrt{10}}$. Thus, $CD = 2\sqrt{15 + 10\sqrt{10}}$.

We are looking for $CE^2$ + $DE^2$ which is also $(CE + DE)^2 - 2 \cdot CE \cdot DE$.

Because $CE + DE = CD = 2\sqrt{15 + 10\sqrt{10}}$, $(CE + DE)^2 = CD^2=4(15 + 10\sqrt{10}) = 60 + 40\sqrt{10}$.

By Power of a Point, $CE \cdot DE = AE \cdot BE = 2\sqrt5\cdot(10\sqrt2 - 2\sqrt5) = 20\sqrt{10} - 20$, so $2 \cdot CE \cdot DE = 40\sqrt{10} - 40$.

Finally, $CE^2 + DE^2 = (CE+ED)^2-2\cdot CE \cdot DE=(60 + 40\sqrt{10}) - (40\sqrt{10} - 40) = \boxed{\textbf{(E)}\ 100}$.

Practice Problems