Question 131 markMCQSection A

  1. In the given figure, PA is a tangent from an external point P to a circle with centre O. If ∠POB=125°\angle POB = 125°, then ∠APO is equal to : (a) 25° (b) 65° (c) 90° (d) 35°

Circles
Tangent perpendicular to radius

Options

(A)25°25°
(B)65°65°
(C)90°90°
(D)35°35°

Official answer

(d) 35°35°, since OA⊥PAOA \perp PA makes ∠OAP=90°\angle OAP = 90°, and the angle sum of triangle OAP gives ∠APO=180°−90°−55°=35°\angle APO = 180° - 90° - 55° = 35°.

Answer keywords:tangentradiusperpendicularpoint of contact∠OAP = 90°angle sum property35 degreesexternal point

Marking scheme

  1. 1.1 mark for the correct option (d) 35°.
  2. 2.Working expected: ∠OAP=90°\angle OAP = 90° (radius ⊥ tangent); ∠AOP=180°−125°=55°\angle AOP = 180° - 125° = 55°; ∠APO=180°−90°−55°=35°\angle APO = 180° - 90° - 55° = 35°.
  3. 3.No marks for merely writing 35° with a contradictory reason; accept the answer if radius–tangent perpendicularity is used correctly.

Hint

The radius meets the tangent at 90° (∠OAP=90°\angle OAP = 90°); find ∠AOP using the straight line at O, then use angle sum of triangle OAP.

Quick oral answer

Since the radius OA is perpendicular to the tangent PA, ∠OAP is 90°; ∠AOP is 180° minus 125°, which is 55°, so ∠APO is 180° minus 90° minus 55°, giving 35°.

Common mistakes

  1. 1.Selecting (c) 90°90° by reporting the radius–tangent angle ∠OAP instead of the required ∠APO.
  2. 2.Forgetting that ∠POB and ∠AOP are supplementary and using 125°125° directly inside the triangle, giving a negative or impossible angle.
  3. 3.Assuming triangle OAP is isosceles (OA=PAOA = PA) — OA is a radius and PA is a tangent length, so they are generally unequal.

Analysis and explanation

Tests the tangent–radius perpendicularity theorem to find an unknown angle in a triangle formed by the centre, the point of contact, and the external point.


Concept

  • The tangent to a circle at any point is perpendicular to the radius drawn to that point of contact, so OA⊥PAOA \perp PA and ∠OAP=90°\angle OAP = 90°.
  • A, O, B lie such that ∠AOP is the supplement of the given ∠POB (linear pair on line AOB), giving ∠AOP=180°−125°=55°\angle AOP = 180° - 125° = 55°.
  • In triangle OAP, angle sum gives ∠APO=180°−90°−55°=35°\angle APO = 180° - 90° - 55° = 35°.

Common mistakes

  • Forgetting the radius–tangent right angle and instead assuming ∠OAP is some other value.
  • Adding 125° and 90° directly instead of first finding the supplementary angle ∠AOP.

Interesting facts

The theorem that a tangent is perpendicular to the radius at the point of contact appears as Theorem 10.1 in the NCERT Class 10 Maths textbook (Chapter 10, Circles).

Because OA ⊥ PA, the point A always lies on the circle with OP as diameter — a neat consequence used to construct tangents from an external point.

Question details

Q13 of 53 questions

Board
CBSE
Class
Class 10
Subject
Mathematics (Basic)
Year
2026
Section
Section A
Marks
1
Type
MCQ
Chapter
Circles
Topic
Tangent perpendicular to radius
Bloom's level
apply
Syllabus topic
Unit 4: Geometry — Circles: Tangent to a circle is perpendicular to the radius at the point of contact
Author
CBSE
Reviewed by
Padhantu Academic Team

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Frequently asked questions

Why is ∠OAP\angle OAP equal to 90°90°?

Because A is the point of contact of the tangent PA, and the tangent at any point of a circle is perpendicular to the radius (OA) drawn to that point. This is a standard circle theorem.

How do we get ∠AOP=55°\angle AOP = 55° from ∠POB=125°\angle POB = 125°?

Points A, O and B are arranged so that ∠AOP and ∠POB lie on a straight line at O and are supplementary. Hence ∠AOP=180°−125°=55°\angle AOP = 180° - 125° = 55°.

Could the triangle be solved without the perpendicular property?

No. The right angle at A is what makes triangle OAP solvable with just one more angle; without radius ⊥ tangent you cannot fix any angle of the triangle.