Question 131 markMCQSection A

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°

Circle with centre O; external point P; tangent line PA touching the circle at A, so OA ⊥ PA; AB is a diameter of the ci
Fig. for Q13
Circles
Tangent to a Circle

Options

(A)25°
(B)65°
(C)90°
(D)35°

Official answer

(d) 35° — since ∠OAP=90°\angle OAP = 90° (tangent ⊥ radius) and ∠AOP=180°−125°=55°\angle AOP = 180° - 125° = 55° (linear pair on diameter AOB), so ∠APO=180°−90°−55°=35°\angle APO = 180° - 90° - 55° = 35°.

Answer keywords:tangent to a circleradius perpendicular to tangentlinear pairangle sum property of triangleexternal pointpoint of contact

Marking scheme

  1. 1.1 mark: correctly selecting option (d) 35° with valid reasoning — using ∠OAP=90°\angle OAP = 90° (tangent ⊥ radius) and ∠AOP=180°−125°=55°\angle AOP = 180° - 125° = 55° (linear pair since A, O, B are collinear), then ∠APO=180°−90°−55°=35°\angle APO = 180° - 90° - 55° = 35°.
  2. 2.No partial marks for MCQs — only the final correct option choice earns the mark; working may be shown for self-check but is not separately awarded.

Hint

Tangent ⊥ radius at the point of contact (∠OAP=90°\angle OAP = 90°), and since A, O, B are collinear, ∠AOP+∠POB=180°\angle AOP + \angle POB = 180°.

Quick oral answer

Since PA is a tangent, OA is perpendicular to PA at A, giving a 90° angle. As A, O, B lie on a straight diameter, ∠AOP=180°−125°=55°\angle AOP = 180° - 125° = 55°. In right triangle OAP, ∠APO=180°−90°−55°=35°\angle APO = 180° - 90° - 55° = 35°.

Common mistakes

  1. 1.Assuming ∠AOP equals ∠POB (125°) directly instead of realising A, O, B are collinear and applying the linear pair relation to get 55°.
  2. 2.Forgetting that the tangent is perpendicular to the radius at the point of contact, and hence omitting the 90° angle at A in the triangle OAP.
  3. 3.Arithmetic slip while subtracting from 180° (e.g., writing 180−125−90=−35°180 - 125 - 90 = -35° and ignoring the sign, or mixing up which two angles to subtract).

Analysis and explanation

Uses the tangent–radius perpendicularity property combined with the linear pair on diameter AOB.


Concept

  • Tangent PA ⊥ radius OA at the point of contact, so ∠OAP=90°\angle OAP = 90°.
  • A, O, B are collinear (AB is a diameter), so ∠AOP and ∠POB form a linear pair: ∠AOP+∠POB=180°\angle AOP + \angle POB = 180°.

Key points

  • ∠AOP=180°−125°=55°\angle AOP = 180° - 125° = 55°.
  • In right triangle OAP: ∠OAP+∠AOP+∠APO=180°  ⟹  90°+55°+∠APO=180°  ⟹  ∠APO=35°\angle OAP + \angle AOP + \angle APO = 180° \implies 90° + 55° + \angle APO = 180° \implies \angle APO = 35°.

Common mistakes

  • Treating ∠AOP as equal to ∠POB instead of using the linear pair.
  • Forgetting the tangent ⊥ radius fact and applying the exterior angle theorem to the wrong triangle.

Real-world

  • The tangent-perpendicularity principle is used in engineering drawing, gear design, and satellite orbit calculations wherever a straight edge must just touch a circular boundary without crossing it.

Interesting facts

The tangent to a circle is always perpendicular to the radius at the point of contact, a property first rigorously proved in Euclid's Elements (Book III).

Because OA is perpendicular to PA, angle OAP is always 90 degrees, which makes quadrilateral angle-chasing problems like this solvable using the angle sum property of a quadrilateral (360 degrees).

Question details

Q13 of 56 questions

Board
CBSE
Class
Class 10
Subject
Mathematics (Standard)
Year
2026
Section
Section A
Marks
1
Type
MCQ
Chapter
Circles
Topic
Tangent to a Circle
Bloom's level
apply
Syllabus topic
Unit VI: Geometry — Circles: Tangent from an external point, radius perpendicular to tangent
Author
CBSE
Reviewed by
Padhantu Academic Team

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Frequently asked questions

Why is ∠OAP always 90° in this figure?

Because a tangent to a circle is always perpendicular to the radius drawn to the point of contact — this is a fundamental theorem of the Circles chapter.

How do we know A, O, B are on a straight line?

The figure shows AB passing through the centre O, making AB a diameter, so A, O, B are collinear and ∠AOP, ∠POB form a linear pair summing to 180°.