9 min readConcept sheet2 or 3-mark question; sometimes a case-based question

Problems on tangents: lengths and angles

Most tangent problems use two facts: a right angle at the point of contact and equal tangents from one point. Then apply Pythagoras or the angle sum.

Tangent length√(d² − r²), where d = distance of the point from the centre

Angle between two tangents∠APB = 180° − ∠AOB

Tangents at the ends of a diameterAre parallel

Concentric circlesA chord of the larger circle touching the smaller one is bisected; length = 2√(R² − r²)

Fact, figure and what it gives

Fact	Figure to look for	What it gives
Radius ⟂ tangent	Right triangle OAP	Pythagoras theorem: OP² = OA² + PA²
Equal tangents	PA = PB	ΔPAB is isosceles, so ∠PAB = ∠PBA
Quadrilateral OAPB	Right angles at A and B	∠APB + ∠AOB = 180°
Tangents at the ends of a diameter	Both ⟂ the same diameter	The tangents are parallel
Concentric circles	Chord of the larger circle touching the smaller	The radius to the point of contact bisects the chord

How to draw a tangent problem figure

Draw the circle and mark the centre O.
Draw the tangents and name the points of contact.
Always join the radius to each point of contact and mark the right angle.
Put the same tick marks on equal tangents.
Write every given length and angle on the figure before you start.

Concentric circles

Problem

Two concentric circles have radii 10 cm and 6 cm. Find the length of the chord of the larger circle that touches the smaller circle.

Working

Let the chord AB touch the smaller circle at P. Then OP ⟂ AB (tangent ⟂ radius).
The perpendicular from the centre bisects a chord, so AP = PB.
In right ΔOPA: AP² = OA² − OP² = 10² − 6² = 100 − 36 = 64.
AP = 8 cm, so AB = 2 × 8 = 16 cm.

Answer

16 cm

Using an angle and a radius

Problem

PA and PB are tangents from P to a circle with centre O. ∠APB = 60° and OA = 5 cm. Find OP and PA.

Working

OP bisects ∠APB, so ∠OPA = 30°. Also ∠OAP = 90°.
In right ΔOAP: sin 30° = OA/OP, so 1/2 = 5/OP and OP = 10 cm.
PA² = OP² − OA² = 100 − 25 = 75.
PA = √75 = 5√3 cm.

Answer

OP = 10 cm, PA = 5√3 cm

Angles with the chord of contact

Problem

Tangents TP and TQ are drawn from T to a circle with centre O. ∠PTQ = 40°. Find ∠TPQ and ∠OPQ.

Working

TP = TQ (equal tangents), so ΔTPQ is isosceles and ∠TPQ = ∠TQP.
∠TPQ = (180° − 40°) ÷ 2 = 70°.
∠OPT = 90° (tangent ⟂ radius).
∠OPQ = ∠OPT − ∠TPQ = 90° − 70° = 20°.

Answer

∠TPQ = 70°, ∠OPQ = 20°

In real life

The belt of an atta chakki (flour mill) leaves each pulley along a tangent. Mechanics use the right angle at the point of contact to work out how long the belt must be.

Exam angle

2 marks: “AB is a diameter of a circle. Show that the tangents at A and B are parallel.” Each tangent is ⟂ AB (tangent ⟂ radius), so both make 90° with AB. Co-interior angles add to 180°, so the tangents are parallel.

3 marks: “Two concentric circles have radii 13 cm and 5 cm. Find the length of the chord of the larger circle that touches the smaller circle.” Half chord = √(13² − 5²) = √144 = 12 cm, because the radius to the point of contact bisects the chord. Chord = 24 cm.

3 marks: “PA and PB are tangents from P to a circle with centre O. Prove that ∠APB + ∠AOB = 180°.” ∠OAP = ∠OBP = 90°. Angles of quadrilateral OAPB add to 360°, so ∠APB + ∠AOB = 360° − 180° = 180°.

Remember it

Spot the right angle first: every radius drawn to a point of contact makes 90° with the tangent.