Secants and Products

Two chords $AB$ and $CD$ meet at $X$ inside a circle. Join $A$ to $C$ and $B$ to $D$, making two triangles that share the vertex $X$. Angles in the same segment are equal, and vertically opposite angles are equal. Count how many pairs of equal angles you now have between those two triangles, and say what that proves about them.

When two lines through a point $P$ each meet a circle, the product of the two distances along one line equals the product along the other. The proof is always the same: join the endpoints crosswise, get similar triangles, write the ratio of matching sides, cross-multiply.

Chords $AB$ and $CD$ meet at $X$ inside the circle.

To prove: $AX \times XB = CX \times XD$.

Construction: join $AC$ and $BD$, crosswise, to make $\triangle AXC$ and $\triangle DXB$.

$\angle CAX = \angle BDX$, since $\angle CAB$ and $\angle CDB$ stand on the same arc $CB$ (angles in the same segment). Also $\angle AXC = \angle DXB$ (vertically opposite). Two pairs of equal angles is enough, so

$$\triangle AXC \sim \triangle DXB \quad \text{(equiangular)}$$

Matching sides are in the same ratio. Take the two sides that meet at $X$ in each triangle:

$$\frac{AX}{DX} = \frac{CX}{BX}$$

Cross-multiplying gives $AX \times BX = CX \times DX$. That is the whole theorem.

Now let $P$ be outside the circle, with one secant through $A$ then $B$, and another through $C$ then $D$, where $A$ and $C$ are the near intersections.

To prove: $PA \times PB = PC \times PD$.

Construction: join $AD$ and $CB$, crosswise again, giving $\triangle PAD$ and $\triangle PCB$.

$\angle APD = \angle CPB$, since they are the same angle at $P$. And $ABDC$ is a cyclic quadrilateral, so $\angle PAD = 180° - \angle DAB = \angle DCB = \angle PCB$, using the exterior angle result from Lesson 3. Two pairs of equal angles, so

$$\triangle PAD \sim \triangle PCB \quad \text{(equiangular)}$$

$$\frac{PA}{PC} = \frac{PD}{PB} \quad \Longrightarrow \quad PA \times PB = PC \times PD$$

Note carefully what each product uses: $PA$ is the near length and $PB$ is the whole secant, not the chord $AB$. Multiplying near by chord is the most common error in this topic.

Let $PT$ be a tangent touching at $T$, and let a secant from $P$ cut the circle at $A$ then $B$.

To prove: $PT^2 = PA \times PB$.

Construction: join $TA$ and $TB$, giving $\triangle PTA$ and $\triangle PBT$.

The angle at $P$ is shared. And by the alternate segment theorem from Lesson 4, the tangent-chord angle $\angle PTA$ equals $\angle TBA$, which is $\angle PBT$. So the triangles are equiangular:

$$\frac{PT}{PB} = \frac{PA}{PT} \quad \Longrightarrow \quad PT^2 = PA \times PB$$

This is not a fourth theorem. Slide the second secant until its two intersections merge into one: $PC$ and $PD$ both become $PT$, and $PC \times PD$ becomes $PT^2$. The tangent is the secant's limiting case, and the algebra follows it exactly.

Turning it round, $PT = \sqrt{PA \times PB}$ gives the tangent length from any secant measurement, without ever locating the centre.

Three results, one number. Fix a point $P$ and let a line through it meet the circle at $U$ and $V$. The product $PU \times PV$ turns out to be the same for every line you can draw through $P$. That common value is called the power of $P$.

With centre $O$ and radius $r$, take the line through $P$ and $O$. It meets the circle at distances $|OP - r|$ and $OP + r$ from $P$, so the power is

$$|OP^2 - r^2|$$

Check it against Lesson 4: for a tangent, $PT^2 = OP^2 - r^2$, which is exactly the power. The equal-tangents result is then obvious, since both tangents from $P$ have length equal to the square root of the same number.

The three theorems in this lesson are just the statement "the power does not depend on the line", written for a point inside, a point outside, and the tangent case. You are not asked to use the word power in an exam, but knowing the three are one result stops you memorising three.

Four problems using the three product results. Work each one, then reveal the answer.

1. 1 Chords meet at $X$ with $AX = 8$, $XB = 3$ and $CX = 6$. Find $XD$.

   Show answer
   $8 \times 3 = 6 \times XD$, so $XD = 4$.$4$

2. 2 From $P$, a secant has $PA = 3$ and $PB = 12$. A tangent from $P$ touches at $T$. Find $PT$.

   Show answer
   $PT^2 = 3 \times 12 = 36$.$PT = 6$

3. 3 From $P$, $PC = 5$ and $CD = 7$. A second secant has $PA = 4$. Find the whole secant $PB$.

   Show answer
   $PD = 5 + 7 = 12$, so $4 \times PB = 60$.$PB = 15$

4. 4 A circle has radius $5$ and $P$ is $9$ from the centre. Find the power of $P$, and the tangent length.

   Show answer
   $9^2 - 5^2 = 81 - 25 = 56$, and the tangent length is its square root.power $56$, $PT = 2\sqrt{14}$