Equal Tangents from a Point

\( PA=PB \)
Circle Theorems GCSE

In ΔOAP and ΔOBP, why are PA and PB equal?

Hint (H)
Think of congruence rules.

Explanation

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Statement

If two tangents are drawn to a circle from the same external point, then the lengths of the tangents are equal. Symbolically:

\[ PA = PB \]

where \(PA\) and \(PB\) are tangents from point \(P\) to the circle at points \(A\) and \(B\).

Why it’s true

  • Draw radii \(OA\) and \(OB\) to the tangent points.
  • \(\angle OAP = \angle OBP = 90^\circ\) because a radius is perpendicular to a tangent at the point of contact.
  • Triangles \(\triangle OAP\) and \(\triangle OBP\) are congruent (right angle, common hypotenuse \(OP\), and radius \(OA=OB\)).
  • By congruence, \(PA = PB\).

Recipe (how to use it)

  1. Identify the external point and the two tangents drawn to the circle.
  2. Apply the rule: the tangents from the same external point are equal in length.
  3. Use this equality to solve for unknown lengths in geometry problems.

Spotting it

Look for problems where a circle and tangents from an external point are mentioned. The key phrase is “tangents from the same point”.

Common pairings

  • Circle theorems involving tangents and radii.
  • Right-angled triangles formed by radii and tangents.
  • Geometry proofs requiring equal lengths.

Mini examples

  1. Given: A point \(P\) outside a circle has tangents \(PA\) and \(PB\). If \(PA=8\), then \(PB=8\).
  2. Given: From point \(P\), tangents \(PA\) and \(PB\) touch a circle of radius 5 cm. If \(OP=13\), then \(PA=PB=\sqrt{13^2-5^2} = 12\).

Pitfalls

  • Confusing tangents with secants (tangents touch once, secants cut twice).
  • Forgetting that both tangents from the same point are equal, not just similar.

Exam strategy

  • Always mark the two tangents as equal when drawn from the same point.
  • Use congruent triangles (OAP and OBP) to justify in proofs.
  • Check whether Pythagoras can be applied in the right-angled triangles formed.

Summary

Tangents drawn to a circle from a single external point are equal in length. This is a standard circle theorem and is useful for proving equal lengths and solving geometry problems.

Worked examples

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  1. \( From a point P outside a circle, two tangents PA and PB are drawn. If PA=7 cm, find PB. \)
    1. Tangents from same point are equal.
    2. \( So PB=7 cm. \)
    Answer: 7 cm
  2. \( A circle has radius 5 cm and OP=13 cm, where O is the centre and P is an external point. Find the length of the tangent PA. \)
    1. Use right triangle OAP.
    2. \( OP^2 = OA^2 + AP^2 \)
    3. \( 13^2 = 5^2 + AP^2 \)
    4. \( 169=25+AP^2 \)
    5. \( AP^2=144 \)
    6. \( AP=12 \)
    Answer: 12 cm
  3. Tangents PA and PB are drawn from a point P to a circle with centre O. Show that ΔOAP ≅ ΔOBP.
    1. \( OA=OB (radii) \)
    2. \( OP=OP (common) \)
    3. \( ∠OAP=∠OBP=90° \)
    4. So triangles are congruent by RHS.
    5. \( Thus PA=PB. \)
    Answer: \( PA=PB \)
  4. Two tangents are drawn from a point 10 cm from the centre of a circle of radius 6 cm. Find tangent length.
    1. Right triangle OAP
    2. \( OP^2=OA^2+AP^2 \)
    3. \( 10^2=6^2+AP^2 \)
    4. \( 100=36+AP^2 \)
    5. \( AP^2=64 \)
    6. \( AP=8 \)
    Answer: 8 cm
  5. \( From a point outside a circle, tangents PA and PB are drawn. If PA=9 cm, what is PB? \)
    1. Tangents from same point are equal.
    2. \( So PB=9 cm. \)
    Answer: 9 cm
  6. \( In a circle, P is a point outside with PA and PB as tangents. Prove PA=PB using congruent triangles. \)
    1. \( OA=OB (radii) \)
    2. \( ∠OAP=∠OBP=90° \)
    3. \( OP=OP (common) \)
    4. So ΔOAP≅ΔOBP.
    5. \( Therefore, PA=PB. \)
    Answer: \( PA=PB \)
  7. If a tangent from a point P to a circle is 15 cm and the radius is 9 cm, find OP (distance from P to O).
    1. Right triangle OAP
    2. \( OP^2=OA^2+AP^2 \)
    3. \( OP^2=9^2+15^2=81+225=306 \)
    4. \( OP=√306≈17.49 \)
    Answer: 17.49 cm
  8. From a point outside a circle of radius 12 cm, tangents are drawn. If tangent length is 16 cm, find distance from point to centre.
    1. Right triangle OAP
    2. \( OP^2=OA^2+AP^2 \)
    3. \( OP^2=12^2+16^2=144+256=400 \)
    4. \( OP=20 \)
    Answer: 20 cm
  9. Point P is 25 cm from centre O of a circle of radius 24 cm. Find tangent length.
    1. Right triangle OAP
    2. \( AP^2=OP^2-OA^2=25^2-24^2=625-576=49 \)
    3. \( AP=7 \)
    Answer: 7 cm
  10. From point P, two tangents PA and PB touch a circle at A and B. Which triangle pair is congruent?
    1. Triangles OAP and OBP are congruent by RHS rule.
    2. \( Hence PA=PB. \)
    Answer: ΔOAP≅ΔOBP