Equal Tangents from an External Point (Theorem 10.2)

Theorem 10.2 — Statement

We saw that two tangents can be drawn from an external point, and that they look equal. Now we prove it.

Theorem 10.2: The lengths of the two tangents drawn from an external point to a circle are equal.

If $P$ is an external point and $PQ$, $PR$ are the two tangents (touching the circle at $Q$ and $R$), then $$PQ = PR.$$

Proof of Theorem 10.2

Given: A circle with centre $O$, an external point $P$, and two tangents $PQ$ and $PR$ (points of contact $Q$ and $R$). To prove: $PQ = PR$.

Construction: Join $OP$, $OQ$ and $OR$.

Proof. By Theorem 10.1, a tangent is perpendicular to the radius at the point of contact, so $$\angle OQP = \angle ORP = 90^\circ.$$

Now compare the right triangles $OQP$ and $ORP$:

• $OQ = OR$ (radii of the same circle),
• $OP = OP$ (common hypotenuse),
• $\angle OQP = \angle ORP = 90^\circ$.

By the RHS congruence rule, $\triangle OQP \cong \triangle ORP$.

Hence, by CPCT (corresponding parts of congruent triangles), $$PQ = PR. \qquad \blacksquare$$

A Second Proof, and the Angle-Bisector Property

Pythagoras version. In the right triangles, $PQ^2 = OP^2 - OQ^2$ and $PR^2 = OP^2 - OR^2$. Since $OQ = OR$, we get $PQ^2 = PR^2$, so $PQ = PR$.

Angle-bisector property. The same congruence $\triangle OQP \cong \triangle ORP$ also gives $$\angle OPQ = \angle OPR \quad\text{and}\quad \angle QOP = \angle ROP.$$

So $OP$ bisects the angle $\angle QPR$ between the two tangents, and also bisects the angle $\angle QOR$ at the centre.

Key Point: The line joining the external point to the centre bisects both the angle between the tangents and the angle between the radii to the points of contact.

A Very Useful Angle Relation

Look at the quadrilateral $OQPR$ (centre $O$, contacts $Q$, $R$, external point $P$). Its angles at $Q$ and $R$ are each $90^\circ$. Since the angles of a quadrilateral add up to $360^\circ$, $$\angle QOR + \angle QPR = 360^\circ - 90^\circ - 90^\circ = 180^\circ.$$

Result: The angle between the two tangents from an external point and the angle subtended by the line joining the points of contact at the centre are supplementary: $$\angle QPR + \angle QOR = 180^\circ.$$

[Board Important] So if $\angle POQ = 110^\circ$, then $\angle PTQ = 180^\circ - 110^\circ = 70^\circ$; if the tangents meet at $80^\circ$, then $\angle$ at the centre $= 100^\circ$, and each $\angle OPA = \tfrac{1}{2}(180^\circ - 80^\circ) = 50^\circ$.

Solved Examples

Example 1: Angle between tangents

Two tangents $TP$ and $TQ$ are drawn to a circle with centre $O$ from an external point $T$, and $\angle POQ = 110^\circ$. Find $\angle PTQ$.

Solution:

1. $\angle PTQ$ and $\angle POQ$ are supplementary: $\angle PTQ + \angle POQ = 180^\circ$.
2. $\angle PTQ = 180^\circ - 110^\circ = 70^\circ$.

Final Answer: $\angle PTQ = 70^\circ$.

Takeaway: Angle between tangents $+$ angle at centre $= 180^\circ$.

Example 2: Find the angle at the centre

Tangents $PA$ and $PB$ from a point $P$ to a circle with centre $O$ are inclined to each other at $80^\circ$. Find $\angle POA$.

Solution:

1. $\angle AOB = 180^\circ - \angle APB = 180^\circ - 80^\circ = 100^\circ$.
2. $OP$ bisects $\angle AOB$, so $\angle POA = \tfrac{1}{2}(100^\circ) = 50^\circ$.

Final Answer: $\angle POA = 50^\circ$.

Takeaway: $OP$ bisects the central angle; half of $(180^\circ - \text{angle between tangents})$.

Example 3: Isosceles set-up

From an external point $T$, tangents $TP$ and $TQ$ touch a circle at $P$ and $Q$. If $\angle PTQ = 60^\circ$, find $\angle TPQ$.

Solution:

1. $TP = TQ$ (equal tangents), so triangle $TPQ$ is isosceles with $\angle TPQ = \angle TQP$.
2. Angle sum: $\angle TPQ + \angle TQP + 60^\circ = 180^\circ \Rightarrow 2\angle TPQ = 120^\circ$.
3. $\angle TPQ = 60^\circ$.

Final Answer: $\angle TPQ = 60^\circ$ (the triangle is in fact equilateral).

Takeaway: Equal tangents make triangle $TPQ$ isosceles — use the base angles.