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Two Parallel Tangents of a Circle Meet a Third Tangent at Points P and Q. Prove that Pq Subtends a Right Angle at the Centre. - Mathematics

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Question

Two parallel tangents of a circle meet a third tangent at points P and Q. prove that PQ subtends a right angle at the centre.

Solution

 Join OP, OQ, OA, OB and OC.
In Δ OAP and Δ OCP

OA = OC (Radii of the same circle)
OP = OP (Common)
PA = PC (Tangents from P)
∴ By side – side – side criterion of congruence,
ΔOAP  ≅ Δ OCP (SSS postulate)
The corresponding parts of the congruent triangles are congruent.
⇒ `∠`APO = `∠`CPO     (cpct) ………….(i)
Similarly, we can prove that
∴ Δ OCQ  ≅  ΔOBQ
⇒ `∠` CQO = `∠`BQO ……………(ii)
∴ `∠` APC = 2`∠`CPO and `∠`CQB= 2`∠`CQO
But,
`∠`APC = `∠` CQB  = 180
(Sum of interior angles of a transversal)
∴ 2`∠`CPO + 2`∠`CQO = 180°

⇒ `∠`CPO + `∠`CQO = 90°
Now in ΔPOQ,
`∠`CPO  + `∠`CQO + `∠`POQ = 180°

⇒90° + `∠`POQ  = 180°
∴ `∠` POQ  = 90°

  Is there an error in this question or solution?
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APPEARS IN

 Selina Solution for Concise Mathematics for Class 10 ICSE (2020 (Latest))
Chapter 18: Tangents and Intersecting Chords
Exercise 18 (A) | Q: 15 | Page no. 275
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Two Parallel Tangents of a Circle Meet a Third Tangent at Points P and Q. Prove that Pq Subtends a Right Angle at the Centre. Concept: Tangent Properties - If Two Circles Touch, the Point of Contact Lies on the Straight Line Joining Their Centers.
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