Chapter 10 of 14

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Rajasthan (RBSE)Class 10 Mathematics★ 3–5 marks · RBSE Class 10 Maths board (Unit — Geometry)

Circles

RBSE Class 10 Maths on tangents to a circle — the two key theorems (a tangent is perpendicular to the radius at the point of contact, and the two tangents from an external point are equal) and how to use them in proofs and calculations, with fully solved board-pattern questions.

⏱ 24 min readMedium difficulty✓ Free · NCERT-aligned

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Study Guide

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Key points, glossary and the questions most likely to show up.

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By the end of this chapter you'll be able to…

1Distinguish a non-intersecting line, a secant and a tangent
2State and use: the tangent is perpendicular to the radius at the point of contact
3State and use: tangents from an external point are equal in length
4Find tangent lengths using Pythagoras
5Prove standard tangent results (e.g. opposite sides of a circumscribing quadrilateral)

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Why this chapter matters

A compact, high-return geometry chapter resting on just two tangent theorems. It reliably yields a proof and a short right-triangle calculation, and the 'tangent ⟂ radius' idea makes many problems one-step Pythagoras.

Before you start — revise these

A 5-minute refresher here will save you 30 minutes of confusion below.

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Circles (Class 9)

Chords, and the perpendicular from the centre bisecting a chord.

Circles — RBSE Class 10 (Mathematics)

A line and a circle can miss, cross at two points, or just kiss at one. That single-touch line is a tangent, and the whole of this compact chapter hangs on two elegant facts about tangents — facts that generate clean proofs and neat right-angle calculations.

1. A line and a circle — three cases

Non-intersecting line — no common point.
Secant — a line that cuts the circle at two points (contains a chord).
Tangent — a line touching the circle at exactly one point, the point of contact.

A tangent is the limiting position of a secant as its two intersection points merge.

2. Theorem 1 — tangent ⟂ radius

The tangent at any point of a circle is perpendicular to the radius through the point of contact.

So if $O P$ is the radius to the point of contact $P$ and $X Y$ is the tangent there, then $O P ⊥ X Y$ . This right angle is the workhorse: it turns tangent problems into right-triangle (Pythagoras) problems.

Consequence: the length of a tangent from an external point A to a circle of radius $r$ with centre O is $A P = O A^{2} - r^{2} .$

3. Theorem 2 — equal tangents from an external point

The lengths of the two tangents drawn from an external point to a circle are equal.

If AP and AQ are tangents from A, then $A P = A Q$ . Moreover, they subtend equal angles at the centre and the line from the external point to the centre bisects the angle between the tangents.

These two theorems, plus the earlier "perpendicular from centre bisects a chord," settle almost every question in the chapter.

4. How the proofs usually go

To prove a right angle or a length: use Theorem 1 and Pythagoras in triangle $O P A$ .
To prove two segments equal, or to find a perimeter of a tangential figure: use Theorem 2 (equal tangents).
Classic result: in a quadrilateral circumscribing a circle, the sums of opposite sides are equal, $A B + C D = B C + D A$ — proved directly from equal tangents at the four contact points.

5. Worked idea

Two concentric circles have radii 5 cm and 3 cm. Find the length of the chord of the larger circle that touches the smaller one. The chord is tangent to the inner circle, so the radius (3 cm) to the contact point is perpendicular to it and bisects it. Half-chord $= 5^{2} - 3^{2} = 4$ , so the chord $= 8$ cm.

6. Closing thought

Two theorems — tangent ⟂ radius and equal tangents from an external point — carry the entire chapter. Learn to state them precisely and to spot which one a figure needs. In the RBSE board this reliably yields a proof plus a short calculation, and the right-angle-at-contact idea makes many problems fall out with a single use of Pythagoras.

Key formulas & results

Everything you need to memorise, in one card. Screenshot this for revision.

Tangent ⟂ radius

OP ⟂ tangent at P

Right angle at the point of contact.

Tangent length

AP = √(OA² − r²)

From external point A, radius r.

Equal tangents

AP = AQ

Two tangents from the same external point.

Number of tangents

external: 2, on circle: 1, inside: 0

Depends on the point's position.

Circumscribing quadrilateral

AB + CD = BC + DA

From equal tangents at the four contacts.

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Common mistakes & fixes

These are the exact errors that cost students marks in board exams. Read them once, save yourself the trouble.

WATCH OUT

✗ Assuming the tangent passes through the centre

✓ The RADIUS to the contact point passes through the centre; the tangent is perpendicular to it, not through the centre.

WATCH OUT

✗ Using OA as the tangent length

✓ OA is the distance to the centre (hypotenuse). The tangent length is AP = √(OA² − r²).

WATCH OUT

✗ Forgetting the right angle at the point of contact

✓ Always mark OP ⟂ tangent — it is the key to setting up the right triangle.

WATCH OUT

✗ Miscounting tangents from a point

✓ Two from an external point, exactly one from a point on the circle, none from an interior point.

WATCH OUT

✗ Not using equal tangents for perimeters

✓ In tangential figures, group equal tangent segments to simplify side sums.

NCERT exercises (with solutions)

Every NCERT exercise from this chapter — what it covers and how many questions to expect.

Ex 10.1

Exercise 10.1

Number of tangents; basic tangent length

Questions

Ex 10.2

Exercise 10.2

Tangent theorems, proofs and calculations

Questions

Examples (in-text)

Worked tangent proofs and lengths

Questions

Practice problems

Try each one yourself before tapping "Show solution". Active recall > rereading.

Q1EASY· Concept

How many tangents can be drawn to a circle from a point outside it?

▶ Show solution

Step 1 — From an external point exactly two tangents can be drawn.

✦ Answer: two.

Q2EASY· Tangent length

A tangent from a point 13 cm from the centre touches a circle of radius 5 cm. Find the tangent length.

▶ Show solution

Step 1 — AP = √(13² − 5²) = √(169 − 25) = √144.

✦ Answer: 12 cm.

Q3EASY· Angle

The radius OP meets tangent XY at P. What is ∠OPY?

▶ Show solution

Step 1 — The tangent is perpendicular to the radius at the contact point.

✦ Answer: 90°.

Q4MEDIUM· Equal tangents

From an external point A, tangents AP and AQ are drawn. If ∠PAQ = 50°, find ∠OPA and ∠POQ.

▶ Show solution

Step 1 — ∠OPA = 90° (tangent ⟂ radius).

Step 2 — In quadrilateral OPAQ, angles sum to 360°: ∠POQ = 360 − 90 − 90 − 50 = 130°.

✦ Answer: ∠OPA = 90°, ∠POQ = 130°.

Q5MEDIUM· Chord tangent

Two concentric circles have radii 5 cm and 3 cm. Find the length of the chord of the larger circle that touches the smaller.

▶ Show solution

Step 1 — The chord is tangent to the inner circle; the 3 cm radius is perpendicular and bisects it.

Step 2 — Half-chord = √(5² − 3²) = 4 cm ⇒ chord = 8 cm.

✦ Answer: 8 cm.

Q6MEDIUM· Proof

Prove that the tangents drawn from an external point to a circle are equal in length.

▶ Show solution

Step 1 — Let tangents from A touch the circle (centre O) at P and Q. Join OP, OQ, OA.

Step 2 — In △OPA and △OQA: OP = OQ (radii), ∠OPA = ∠OQA = 90° (tangent ⟂ radius), OA common.

Step 3 — By RHS congruence, △OPA ≅ △OQA ⇒ AP = AQ. ∎

✦ Answer: AP = AQ by RHS congruence.

Q7MEDIUM· Parallelogram

Prove that a parallelogram circumscribing a circle is a rhombus.

▶ Show solution

Step 1 — For a circumscribing quadrilateral, AB + CD = BC + DA (equal tangents).

Step 2 — In a parallelogram AB = CD and BC = DA, so 2AB = 2BC ⇒ AB = BC.

Step 3 — Adjacent sides equal ⇒ all sides equal ⇒ rhombus. ∎

✦ Answer: it is a rhombus.

Q8HARD· Quadrilateral

A quadrilateral ABCD circumscribes a circle. If AB = 6 cm, BC = 7 cm and CD = 4 cm, find DA.

▶ Show solution

Step 1 — Opposite sides: AB + CD = BC + DA.

Step 2 — 6 + 4 = 7 + DA ⇒ DA = 3 cm.

✦ Answer: DA = 3 cm.

Q9HARD· Triangle incircle

A circle touches the sides BC, CA, AB of △ABC at D, E, F. If AF = 6, BD = 8, CE = 4, find the perimeter.

▶ Show solution

Step 1 — Equal tangents: AF = AE = 6, BD = BF = 8, CE = CD = 4.

Step 2 — AB = AF+FB = 14, BC = BD+DC = 12, CA = CE+EA = 10.

Step 3 — Perimeter = 14 + 12 + 10 = 36 cm.

✦ Answer: 36 cm.

Q10HARD· Right angle proof

Prove that the tangent at any point of a circle is perpendicular to the radius through the point of contact.

▶ Show solution

Step 1 — Let XY be the tangent at P, centre O. Take any other point Q on XY.

Step 2 — Q lies outside the circle (only P is on it), so OQ > OP.

Step 3 — Thus OP is the shortest distance from O to XY, and the shortest distance is the perpendicular ⇒ OP ⟂ XY. ∎

✦ Answer: OP is the shortest segment, hence perpendicular.

5-minute revision

The whole chapter, distilled. Read this the night before the exam.

•A tangent touches a circle at exactly one point; a secant cuts it at two.
•Theorem 1: tangent ⟂ radius at the point of contact.
•Tangent length from external point: AP = √(OA² − r²).
•Theorem 2: two tangents from an external point are equal.
•The line from an external point to the centre bisects the angle between the tangents.
•Circumscribing quadrilateral: AB + CD = BC + DA.
•Two tangents from an external point subtend equal angles at the centre.

Rajasthan (RBSE) marks blueprint

Where the marks come from in this chapter — so you can plan your prep.

Typical chapter weightage: 3–5 marks

Question type	Marks each	Typical count	What it tests
MCQ / very short	1	1	Number of tangents; right angle; tangent length
Short answer	2–3	1	Equal tangents; chord tangent to inner circle
Long answer / proof	3–4	1	Tangent theorem proof or circumscribing figure

Prep strategy

Memorise both theorems and be able to prove each
Mark the right angle at the point of contact in every figure
Use equal tangents to simplify perimeter and side problems
Practise the concentric-circle chord and incircle problems

Where this shows up in the real world

This chapter isn't just an exam topic — it lives in the world around you.

Belt and pulley systems

The straight portions of a belt around pulleys are tangents to the pulley circles.

Road and rail design

Curves join straight sections tangentially for smooth transitions.

Gears and cams

Contact points between meshing wheels behave like tangent contacts.

Optics

A light ray grazing a lens or mirror edge meets it along a tangent.

Exam strategy

Battle-tested tips from teachers and toppers for this chapter.

State the theorem you use by name in every proof.
Always mark the 90° angle at the point of contact.
For tangent lengths, set up the right triangle O–P–A and use Pythagoras.
In circumscribing-figure problems, label equal tangent segments and add.
Use RHS congruence to prove the equal-tangents theorem cleanly.

Going beyond the textbook

For olympiad aspirants and curious learners — topics that build on this chapter.

The tangent–chord angle (alternate segment theorem).
Power of a point and the tangent–secant length relation.
The incircle, excircles and their tangent lengths (s − a formulas).
Common tangents to two circles (direct and transverse).

Where else this chapter is tested

CBSE board isn't the only one — other exams test this chapter too.

RBSE Class 10 Board (BSER Ajmer)High — a tangent proof plus a short calculation almost every year

NTSE / state scholarshipMedium — circle geometry MCQs

JEE FoundationMedium — tangents recur in coordinate circle geometry

Maths Olympiad (IMO/NMTC)High — circle theorems are central

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Questions students ask

The real ones — pulled from the Q&A community and tutor sessions.

Yes — RBSE (BSER, Ajmer) prescribes the NCERT Mathematics textbook; the chapters match the national syllabus while RBSE sets its own exam pattern.

(1) The tangent at a point is perpendicular to the radius at that point; (2) the two tangents from an external point are equal in length. Almost every question uses one of these.

Two if the point is outside the circle, exactly one if it lies on the circle, and none if it is inside.

Use the right angle at the contact point: tangent length = √(OA² − r²), where OA is the distance to the centre and r is the radius.

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