In the given figure, PA is a tangent from an external point P to a circle with centre O. If $\angle POB = 125°$ , 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°

(D)35°

Official Answer

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

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

Marking Scheme

11 mark: correctly selecting option (d) 35° with valid reasoning — using $\angle OAP = 90°$ (tangent ⊥ radius) and $\angle AOP = 180° - 125° = 55°$ (linear pair since A, O, B are collinear), then $\angle APO = 180° - 90° - 55° = 35°$ .
2No 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 ( $\angle OAP = 90°$ ), and since A, O, B are collinear, $\angle AOP + \angle POB = 180°$ .

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, $\angle AOP = 180° - 125° = 55°$ . In right triangle OAP, $\angle APO = 180° - 90° - 55° = 35°$ .

Analysis & 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 $\angle OAP = 90°$ .
A, O, B are collinear (AB is a diameter), so ∠AOP and ∠POB form a linear pair: $\angle AOP + \angle POB = 180°$ .

Key points

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

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.

Common Mistakes

1Assuming ∠AOP equals ∠POB (125°) directly instead of realising A, O, B are collinear and applying the linear pair relation to get 55°.
2Forgetting 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.
3Arithmetic slip while subtracting from 180° (e.g., writing $180 - 125 - 90 = -35°$ and ignoring the sign, or mixing up which two angles to subtract).

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).

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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°.