Two Crossing Chords Always Cut Each Other Into Equal Products

Draw any line through a fixed point $P$ so that it meets a circle at $A$ and $B$. **The product $PA\times PB$ is the same for every such line. This single idea gives three results:

- Chords crossing inside the circle
- Secants meeting outside the circle
- A tangent and a secant from an outside point**

This part covers each of them, then uses equal tangents in problems on the incircle and circumcircle of a triangle.

**If chords $AB$ and $CD$ intersect at $P$ inside a circle, then $PA\times PB=PC\times PD$.

Why.** Triangles $APC$ and $DPB$ have vertically opposite angles at $P$, and $\angle CAB=\angle CDB$ because they stand on the same arc. So they are similar, and $\frac{PA}{PD}=\frac{PC}{PB}$.

Worked example 1. $PA=4$ cm, $PB=6$ cm and $PC=3$ cm.

$$4\times 6=3\times PDPD=8\text{ cm}$$

Worked example 2. Chord $AB$ is $10$ cm long and chord $CD$ cuts it at $P$ with $PC=3$ cm and $PD=7$ cm. Let $AP=x$.

$$x(10-x)=21x^2-10x+21=0x=3\text{ or }7$$

So $P$ divides $AB$ into $3$ cm and $7$ cm.

Worked example 3. A circle has radius $13$ cm and a chord is $5$ cm from the centre. The diameter perpendicular to it is split into $13-5=8$ and $13+5=18$ cm, so the half-chord $h$ satisfies

$$h^2=8\times 18=144h=12\text{chord}=24\text{ cm}$$

An everyday example. Two wooden bars crossing inside a round window frame follow this rule.

The substance. It is the products that match, not the sums — the two chords can have very different lengths.

**If two secants from an external point $P$ cut the circle at $A,B$ and $C,D$, then $PA\times PB=PC\times PD$, where each length is measured from $P$.

Worked example 1.** $PA=4$ cm and $AB=5$ cm, so $PB=9$ cm. On the other secant $PC=3$ cm.

$$4\times 9=3\times PDPD=12\text{ cm}CD=12-3=9\text{ cm}$$

Worked example 2. $PA=6$ cm, $AB=4$ cm and $CD=7$ cm. Find $PC=x$.

$$6\times 10=x(x+7)x^2+7x-60=0(x+12)(x-5)=0x=5\text{ cm}$$

The negative root is rejected because a length cannot be negative.

An everyday example. Two straight roads from a village crossing a circular park obey the same product rule.

The trap. **Use the whole distance $PB$ to the far point**, not the chord length $AB$.

**If a tangent from $P$ touches the circle at $T$ and a secant from $P$ cuts it at $A$ and $B$, then $P{T}^2=PA\times PB$.

Why.** Imagine the secant turning until $A$ and $B$ move together into $T$. The product $PA\times PB$ stays constant, and becomes $PT\times PT$.

Worked example 1. $PA=4$ cm and $PB=9$ cm.

$$P{T}^2=36PT=6\text{ cm}$$

Worked example 2. $PT=8$ cm and $PA=4$ cm.

$$64=4\times PBPB=16\text{ cm}AB=12\text{ cm}$$

Worked example 3 — through the centre. A circle has radius $5$ cm and $P$ is $13$ cm from the centre. The secant through the centre gives $PA=8$ and $PB=18$.

$$P{T}^2=8\times 18=144PT=12\text{ cm}$$

This matches Pythagoras: $\sqrt{13^2-5^2}=12$.

An everyday example. A string pulled tight from a nail to just touch a round bangle is a tangent whose length can be found from any straight cut through the bangle.

The link. The tangent-secant rule is the limiting case of the external chords theorem.

Use equal tangents from each vertex to the incircle, write the sides as sums of tangent lengths, and combine with the circle theorems for chords and the circumcircle.

Worked example 1 — incircle. $△ABC$ has $AB=8$, $BC=10$ and $CA=12$ cm. The incircle touches the sides. Let the tangent lengths from $A$, $B$, $C$ be $x$, $y$, $z$.

$$x+y=8y+z=10x+z=12x+y+z=15$$
$$x=5y=3z=7$$

So the incircle touches $BC$ at a point $3$ cm from $B$ and $7$ cm from $C$.

Worked example 2 — right triangle. Legs $6$ and $8$ cm, hypotenuse $10$ cm.

$$\text{inradius}=\frac{6+8-10}{2}=2\text{ cm}\text{circumradius}=\frac{10}{2}=5\text{ cm}$$

The circumradius is half the hypotenuse because the hypotenuse is a diameter.

An everyday example. A circular badge fitted exactly inside a triangular school pennant touches all three sides, with equal tangent lengths from each corner.

The substance. Each tangent length equals the semi-perimeter minus the opposite side: $x=15-10=5$.