Power of a point

Draw two chords inside a circle and let them cross. Measure the four pieces they get cut into, and something surprising happens every single time: the product of the two pieces of one chord always equals the product of the two pieces of the other. Move the intersection point outside the circle, so the chords become secants meeting beyond it, and a matching rule appears, this time using the full distances from that outside point. Let one of those lines shrink into a tangent, just grazing the circle at a single point, and the rule survives in its most compact form of all, with the tangent length squared. This whole family of results is nicknamed power of a point, and it turns problems that would otherwise need a full similar-triangles argument into one line of algebra. This article rebuilds the three relations from the similar triangles that actually produce them, works through the same tangent-secant and intersecting-chords examples you have already seen, and adds a mixed practice problem to sharpen the "which formula fits this picture" instinct that examiners like to test.

The core insight

Whenever two lines through a fixed point cross a circle, the products of the intercepted-segment lengths on each line are equal, a single similar-triangles fact that yields the intersecting-chords, secant-secant, and tangent-secant formulas as three faces of one idea.

The core idea

Why a "power" at all. For a point $P$ and a circle, every line through $P$ that cuts the circle at two points $A$ and $B$ gives the same signed product $PA \cdot PB$ , no matter which such line you draw. That constant value is called the power of the point $P$ with respect to the circle. You don't need that name for the syllabus, but it explains why three seemingly different pictures all reduce to one number.

Case 1 — two chords meeting inside the circle. If chords $AB$ and $CD$ meet at an interior point $X$ , the two parts of each chord multiply to the same value:

AX \cdot XB = CX \cdot XD

where: $AX, XB$ are the two segments $X$ splits chord $AB$ into, and $CX, XD$ are the two segments it splits chord $CD$ into.

Case 2 — two secants meeting outside the circle. If two lines through an external point $P$ each cut the circle twice, at $A,B$ and at $C,D$ , then the whole distances from $P$ satisfy:

PA \cdot PB = PC \cdot PD

where: $PA, PB$ are the distances from $P$ to the near and far crossing points of the first secant, and $PC, PD$ are the same for the second secant.

Case 3 — a tangent and a secant meeting outside the circle. If $PT$ touches the circle at $T$ and a secant from the same point $P$ cuts the circle at $A$ and $B$ , then:

PT^2 = PA \cdot PB

where: $PT$ is the tangent length from $P$ to the point of contact $T$ . Case 3 is what happens to Case 2 when the two crossing points of one secant slide together and merge into a single point of tangency: $C$ and $D$ both become $T$ , so $PC \cdot PD$ turns into $PT \cdot PT = PT^2$ . Seeing the three cases as one relation, viewed from inside, from outside with two full secants, and from outside with one secant shrunk to a tangent, means you only need to memorise the underlying similar-triangle idea once.

Figure

Three circle diagrams showing chords meeting inside, secants meeting outside, and a tangent with a secant meeting outside, each with matching segment products

Derivation ASimilar triangles behind the intersecting-secants relation

Show, starting only from angles in the same segment and the angle-angle similarity test, that for two secants drawn from an external point $P$ meeting the circle at $A, B$ and at $C, D$ respectively, the products of the whole distances from $P$ satisfy $PA \cdot PB = PC \cdot PD$ , and confirm that shrinking one secant into a tangent recovers $PT^2 = PA \cdot PB$ as the limiting case.

Step 1. Draw the two secants from $P$ , meeting the circle at $A$ then $B$ on one line, and $C$ then $D$ on the other (with $A, C$ the nearer points). Join $AC$ and $BD$ to form triangles $\triangle PAC$ and $\triangle PDB$ .

Step 2. Angle $P$ is common to both triangles, since both are formed by the same two secant lines through $P$ .

Step 3. $ACDB$ is a cyclic quadrilateral (all four points lie on the circle), so its exterior angle at $A$ , namely $\angle PAC$ , equals the interior opposite angle $\angle BDC$ , which is the same angle as $\angle PDB$ (the exterior angle of a cyclic quadrilateral equals the interior opposite angle).

Step 4. With $\angle P$ common and $\angle PAC = \angle PDB$ , triangles $\triangle PAC$ and $\triangle PDB$ are similar by the angle-angle test.

Step 5. Similar triangles give proportional corresponding sides: $\dfrac{PA}{PD} = \dfrac{PC}{PB}$ . Cross-multiplying gives $PA \cdot PB = PC \cdot PD$ , which is exactly the secant-secant relation of Case 2.

Step 6. Now let the secant through $C, D$ rotate toward the circle until it just touches it, so $C$ and $D$ merge into a single point of tangency $T$ . The relation $PA \cdot PB = PC \cdot PD$ becomes $PA \cdot PB = PT \cdot PT = PT^2$ , which is the tangent-secant relation of Case 3, confirming it is not a separate rule but the boundary case of the secant-secant one. The intersecting-chords relation of Case 1 follows the same similar-triangles argument with $P$ moved inside the circle, where the exterior-angle step is replaced by the equal-angles-in-the-same-segment fact instead.

Worked example

Two chords $AB$ and $CD$ of a circle intersect inside the circle at point $X$ , with $AX = 6\ \text{cm}$ , $XB = 4\ \text{cm}$ , $CX = 8\ \text{cm}$ . From an external point $P$ on the same circle's tangent line, a tangent $PT = 12\ \text{cm}$ is drawn, and a secant from $P$ cuts the circle at $A'$ (nearer) with $PA' = 8\ \text{cm}$ .

Step 1. Find $XD$ using the intersecting-chords relation: $AX \cdot XB = CX \cdot XD$ , so $6 \times 4 = 8 \times XD$ , giving $24 = 8 \times XD$ .

Step 2. Solve for $XD$ : $XD = \dfrac{24}{8} = 3\ \text{cm}$ .

Step 3. Find $PB'$ using the tangent-secant relation: $PT^2 = PA' \cdot PB'$ , so $12^2 = 8 \times PB'$ , giving $144 = 8 \times PB'$ .

Step 4. Solve for $PB'$ : $PB' = \dfrac{144}{8} = 18\ \text{cm}$ .

Step 5. Find the far chord segment $A'B'$ : since $A'$ lies between $P$ and $B'$ on the secant line, $A'B' = PB' - PA' = 18 - 8 = 10\ \text{cm}$ . The two independent unknowns work out to $XD = 3\ \text{cm}$ and $A'B' = 10\ \text{cm}$ .

Competitive stretch

Sanity-check the tangent case against equal-tangents-from-a-point: if the secant $PA'B'$ were instead rotated until it too became a tangent (so $A'$ and $B'$ merge), the relation $PT^2 = PA' \cdot PB'$ collapses to $PT^2 = PA'^2$ , giving $PT = PA'$ — exactly the familiar fact that two tangents drawn from the same external point are equal in length. This cross-check confirms the tangent-secant formula is consistent with tangent-length properties learned separately.

Common traps

Confusing the internal-chords formula ( $AX \cdot XB = CX \cdot XD$ , using split parts of each chord) with the external-secant formula ( $PA \cdot PB$ , using whole distances from the external point) — internal cases split segments on both sides of the intersection, external cases measure from the outside point all the way to each crossing.
Forgetting to square the tangent length: writing $PT = PA \cdot PB$ instead of the correct $PT^2 = PA \cdot PB$ .
Swapping near and far points on a secant, leading to $AB = PA - PB$ instead of the correct $PB - PA$ (far minus near).
Applying a formula to a configuration it does not match — always check first whether the intersection point is inside the circle (chords), or outside with two secants, or outside with a tangent and a secant, before picking which relation to use.

Ask the tutor: Stuck deciding whether your diagram needs the chords, secants, or tangent-secant version of this rule? Ask the AI tutor to walk through the configuration with you.

Why does the tangent case use a squared term while the others don't

Because a tangent only touches the circle once, so its "two" intersection points, which would normally be a near point and a far point on a secant, have collapsed into the single point of tangency $T$ . The general secant-secant rule $PA \cdot PB = PC \cdot PD$ still applies, but with $C = D = T$ it becomes $PT \cdot PT$ , which is just $PT^2$ . It's the same rule, only with both factors turning out equal.

Do I need to remember three separate formulas for the exam

You need to be able to state and apply all three, but you don't need to memorise them as unrelated facts. All three come from one similar-triangles argument, and the tangent-secant case is simply the secant-secant case with one line's two crossing points pulled together into one. If you remember the similar-triangles picture, the formula for whichever configuration you're given can be reconstructed rather than recalled from nowhere.

How do I tell from a diagram which of the three cases applies

Look at where the two lines cross relative to the circle. If the crossing point sits inside the circle and both lines are full chords, use the split-segment chords formula. If the crossing point sits outside the circle and both lines cut through it as secants, use the whole-distance secant-secant formula. If one of the two lines only touches the circle at a single point instead of crossing it, that line is a tangent, and you square its length in the tangent-secant formula.

Power of a point

The core idea

Worked example

Competitive stretch

Common traps

Before this

Where this leads