Theorem 10.2 - The lengths of tangents drawn from n external point to circle are equal..jpg

2 Theorem 10.2 - from 1 and 2 PQ  = PR Hence Proved 2.jpg
3 Theorem 10.2 - from 1 and 2 PQ  = PR Hence Proved3 .jpg

4 Theorem 10.2 - from 1 and 2 PQ  = PR Hence Proved 4.jpg 5 Theorem 10.2 - from 1 and 2 PQ  = PR Hence Proved 6.jpg

  1. Chapter 10 Class 10 Circles
  2. Serial order wise
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Theorem 10.2 (Method 1) The lengths of tangents drawn from an external point to a circle are equal. Given: Let circle be with centre O and P be a point outside circle PQ and PR are two tangents to circle intersecting at point Q and R respectively To prove: Lengths of tangents are equal i.e. PQ = PR Construction: Join OQ , OR and OP Proof: As PQ is a tangent OQ ⊥ PQ So, ∠ OQP = 90° Hence Δ OQP is right triangle Similarly, PR is a tangent & OR ⊥ PR So, ∠ ORP = 90° Hence Δ ORP is right triangle Using Pythagoras theorem (Hypotenuse)2 = (Height)2 + (Base)2 From (1) and (2) PQ2 = PR2 PQ = PR Hence proved Theorem 10.2 (Method 2) The lengths of tangents drawn from an external point to a circle are equal. Given: Let circle be with centre O and P be a point outside circle PQ and PR are two tangents to circle intersecting at point Q and R respectively To prove: Lengths of tangents are equal i.e. PQ = PR Construction: Join OQ , OR and OP Proof: As PQ is a tangent OQ ⊥ PQ So, ∠ OQP = 90° Similarly, PR is a tangent & OR ⊥ PR So, ∠ ORP = 90° In Δ OQP and Δ ORP ∠ OQP = ∠ ORP OP = OP OQ = OR ∴ Δ OQP = Δ ORP Hence, PQ = PR Hence both tangents from external point are equal in length Also, ∠ OPQ = ∠ OPR Hence, OP is the angle bisector of ∠ QPR

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Davneet Singh
Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 8 years. He provides courses for Maths and Science at Teachoo. You can check his NCERT Solutions from Class 6 to 12.