Theorem: tangents from an external point are equal
The lengths of tangents drawn from an external point to a circle are equal. Proof: the two right triangles formed are congruent by RHS.
The theorem
Theorem: The lengths of tangents drawn from an external point to a circle are equal. Given: a circle with centre O, a point P outside it, and two tangents PQ and PR touching the circle at Q and R. To prove: PQ = PR. A second way to see it: by Pythagoras theorem, PQ² = OP² − OQ² and PR² = OP² − OR². Since OQ = OR, PQ = PR.
How to draw the figure for the proof
- Draw a circle and mark its centre O.
- Mark a point P outside the circle.
- Draw two tangents from P touching the circle at Q and R.
- Join OQ, OR and OP.
- Mark the right angles at Q and R.
Proof (RHS congruence)
- 1Join OP, OQ and OR.
- 2∠OQP = ∠ORP = 90°, since a tangent is perpendicular to the radius at the point of contact.
- 3In right triangles OQP and ORP: OQ = OR (radii of the same circle) and OP = OP (common hypotenuse).
- 4★So ΔOQP ≅ ΔORP by the RHS congruence rule.
- 5Therefore PQ = PR (CPCT, corresponding parts of congruent triangles). Hence proved.
- 6Also ∠OPQ = ∠OPR (CPCT), so OP bisects the angle between the two tangents.
★ key step
Finding angles
From a point P, two tangents PA and PB are drawn to a circle with centre O. If ∠APB = 70°, find ∠AOB, ∠OPA and ∠POA.
- ∠OAP = ∠OBP = 90° (tangent ⟂ radius).
- Angles of quadrilateral OAPB add to 360°: ∠AOB = 360° − 90° − 90° − 70° = 110°.
- OP bisects ∠APB, so ∠OPA = 70° ÷ 2 = 35°.
- In ΔOAP: ∠POA = 180° − 90° − 35° = 55°.
∠AOB = 110°, ∠OPA = 35°, ∠POA = 55°.
Word problem: two paths to a fountain
Two straight paths from a garden gate P touch a circular fountain at A and B. The fountain has radius 5 m. PA = 12 m. Find PB and the distance of the gate from the centre O.
- Tangents from an external point are equal, so PB = PA = 12 m.
- ∠OAP = 90° (tangent ⟂ radius).
- By Pythagoras theorem, OP² = OA² + PA² = 25 + 144 = 169.
- OP = √169 = 13 m.
PB = 12 m; the gate is 13 m from the centre.
Pull a rope tight round a round well from one point outside it: the two straight parts that leave the well are equal. Builders use this to place two equal paths to a round fountain.
1 marks: “From an external point P, tangents PA and PB are drawn to a circle. If PA = 7 cm, find PB.” PB = 7 cm, because tangents from an external point are equal.
2 marks: “PA and PB are tangents from P to a circle with centre O. If ∠AOB = 130°, find ∠APB.” ∠OAP = ∠OBP = 90°. In quadrilateral OAPB, ∠APB = 360° − 180° − 130° = 50°.
3 marks: “Prove that the lengths of tangents drawn from an external point to a circle are equal.” Figure; join OP, OQ, OR. ∠OQP = ∠ORP = 90°, OQ = OR, OP common. ΔOQP ≅ ΔORP (RHS), so PQ = PR (CPCT).
Two tangents from one point are twins: equal in length, and OP cuts the angle between them in half.