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Chapter 10 · Question 3

State and prove Theorem 10.2: The lengths of tangents drawn from an external point to a circle are equal.

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Q3

State and prove Theorem 10.2: The lengths of tangents drawn from an external point to a circle are equal.

Answer Revealed
Direct Answer:
Statement: The lengths of tangents drawn from an external point to a circle are equal. Proof: Let PP be an external point and PQPQ, PRPR be tangents from PP to a circle with centre OO. Join OPOP, OQOQ, OROR. By Theorem 10.1, ∠OQP=∠ORP=90∘\angle OQP = \angle ORP = 90^\circ. In right triangles OQPOQP and ORPORP, OQ=OROQ = OR (radii) and OP=OPOP = OP (common). By RHS congruence, △OQP≅△ORP\triangle OQP \cong \triangle ORP. By CPCT, PQ=PRPQ = PR. Remarks: ∠OPQ=∠OPR\angle OPQ = \angle OPR, so OPOP is the angle bisector of ∠QPR\angle QPR. Also, using Pythagoras: PQ2=OP2−OQ2=OP2−OR2=PR2PQ^2 = OP^2 - OQ^2 = OP^2 - OR^2 = PR^2, giving PQ=PRPQ = PR.

Simple Explanation

If you draw two tangents from a point outside a circle, both tangent segments are equal in length. It's like the point is connected to the circle by two equally long invisible strings that just touch the circle.

Exam-Ready Structure

Theorem 10.2: The lengths of tangents drawn from an external point to a circle are equal. Given: A circle with centre OO, a point PP outside the circle, and two tangents PQPQ and PRPR from PP touching the circle at QQ and RR. To Prove: PQ=PRPQ = PR. Construction: Join OPOP, OQOQ, and OROR. Proof: ∠OQP=90∘\angle OQP = 90^\circ and ∠ORP=90∘\angle ORP = 90^\circ (by Theorem 10.1, radius is perpendicular to tangent). In right-angled triangles OQPOQP and ORPORP: OQ=OROQ = OR (radii of the same circle), OP=OPOP = OP (common hypotenuse). Therefore, △OQP≅△ORP\triangle OQP \cong \triangle ORP (by RHS congruence rule). By CPCT (Corresponding Parts of Congruent Triangles), PQ=PRPQ = PR. Hence, the lengths of the two tangents are equal. Remarks: (1) ∠OPQ=∠OPR\angle OPQ = \angle OPR, meaning OPOP is the angle bisector of ∠QPR\angle QPR — so the centre lies on the bisector of the angle between the two tangents. (2) Alternative proof using Pythagoras: PQ2=OP2−OQ2=OP2−OR2=PR2PQ^2 = OP^2 - OQ^2 = OP^2 - OR^2 = PR^2, so PQ=PRPQ = PR.

Key Points

  • Two tangents from an external point P to a circle are equal in length
  • Proof: RHS congruence of triangles OQP and ORP
  • OQ = OR (radii), OP = OP (common), angles at Q and R are 90°
  • OP bisects the angle between the two tangents
  • Alternative: Pythagoras theorem proof also works

Related Questions

Q2

State and prove Theorem 10.1: The tangent at any point of a circle is perpendicular to the radius through the point of contact.

Q4

How many tangents can be drawn from a point to a circle when the point is (i) inside the circle, (ii) on the circle, and (iii) outside the circle? Explain with reasoning.

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