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Andhra Pradesh (BIEAP)Class 9 Mathematics★ AP Board (BSEAP) Class 9 Mathematics · ~8 marks · Board: AP Board

Circles

Complete CIRCLE THEOREMS for AP Class 9 — angle at centre = 2× angle at circumference. Angle in a semicircle = 90°. Equal chords are equidistant from centre. Perpendicular from centre to chord bisects it. CYCLIC QUADRILATERALS. Worked proofs and examples.

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1Define: circle, chord, arc, sector, segment, diameter, circumference
2State and apply: angle at centre = 2× angle at circumference
3State and apply: angle in a semicircle = 90°
4State and apply: perpendicular from centre bisects chord (and use Pythagoras to find radius)
5State and apply: equal chords are equidistant from centre
6State and apply: opposite angles of a cyclic quadrilateral sum to 180°
7State and apply: angles in the same segment are equal

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Why this chapter matters

Circles is a proof-heavy geometry chapter that rewards students who learn theorem statements precisely. The angle at centre = twice angle at circumference theorem, the angle in a semicircle = 90° theorem, and the cyclic quadrilateral (opposite angles sum to 180°) are the three most-tested theorems. Numerical problems involving these theorems appear in 3-5 mark questions. The perpendicular from centre to chord bisects it generates Pythagoras-based radius calculations (2-3 marks). Knowing theorem statements word-for-word and being able to write a step-by-step proof is essential for full marks.

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 Mathematics

"The circle is the most PERFECT shape in geometry — every point on its circumference is the SAME distance from its centre."

1. Key Definitions

Term	Definition
Circle	Set of all points in a plane at a FIXED DISTANCE (radius) from a FIXED POINT (centre).
Radius	Line segment joining centre to any point on the circle.
Chord	Line segment joining ANY TWO points on the circle.
Diameter	LONGEST chord — passes through centre. = 2 × radius.
Arc	A PART of the circumference.
Sector	Region between two radii and the included arc.
Segment	Region between a chord and the arc.
Circumfere	Total distance around the circle = 2πr.

2. Theorem 1 — Angle at Centre

The angle subtended by an arc at the CENTRE is TWICE the angle subtended by the SAME arc at ANY POINT on the REMAINING part of the circle.

∠AOB (at centre) = 2 × ∠APB (at circumference), where both are subtended by arc AB.

Worked Example

In a circle, chord AB subtends ∠AOB = 80° at the centre. Find ∠APB where P is a point on the major arc AB. ∠APB = ½ × ∠AOB = ½ × 80° = 40°.

3. Theorem 2 — Angle in a Semicircle

The angle subtended by a DIAMETER at any point on the circle is a RIGHT ANGLE (90°).

This is a SPECIAL CASE of Theorem 1: The angle at centre (by diameter) = 180°. So angle at circumference = ½ × 180° = 90°. 'If AB is a diameter and P is any point on the circle, then ∠APB = 90°. This is one of the MOST frequently used theorems.'

4. Theorem 3 — Perpendicular from Centre to Chord

The perpendicular from the CENTRE of a circle to a CHORD BISECTS the chord.

If OM ⟂ AB, then AM = MB. Converse: The line joining the CENTRE to the MIDPOINT of a chord is PERPENDICULAR to the chord.

Worked Example

A chord of length 16 cm is at a distance of 6 cm from the centre of a circle. Find the radius. Let the chord be AB = 16 cm. Perpendicular OM = 6 cm. AM = ½AB = 8 cm. In right triangle OAM: OA² = OM² + AM² → r² = 6² + 8² = 36 + 64 = 100 → r = 10 cm.

5. Theorem 4 — Equal Chords and Their Distances

Equal chords of a circle are EQUIDISTANT from the centre. Converse: Chords equidistant from the centre are EQUAL.

6. Theorem 5 — Cyclic Quadrilateral

The sum of either pair of OPPOSITE angles of a CYCLIC QUADRILATERAL is 180°.

A cyclic quadrilateral is one whose ALL FOUR VERTICES lie on a circle. If ABCD is cyclic: ∠A + ∠C = 180°. ∠B + ∠D = 180°.

Converse: If the sum of a pair of opposite angles of a quadrilateral is 180°, then the quadrilateral is CYCLIC.

Worked Example

In a cyclic quadrilateral ABCD, ∠A = (2x+10)° and ∠C = (3x−20)°. Find x and the angles. Since ∠A + ∠C = 180°: (2x+10) + (3x−20) = 180 → 5x − 10 = 180 → 5x = 190 → x = 38. ∠A = 2(38)+10 = 86°. ∠C = 3(38)−20 = 94°. Check: 86+94 = 180 ✓.

7. Theorem 6 — Angles in the Same Segment

Angles in the SAME SEGMENT of a circle are EQUAL. If two angles are subtended by the same chord (or arc) and lie on the SAME side of the chord, they are equal.

8. Common Mistakes to Avoid

'Angle in a semicircle is 180°' — NO. The angle at the CENTRE (the diameter is a straight angle) is 180°. The angle at the CIRCUMFERENCE is 90°.
Confusing chord and arc — A chord is a LINE SEGMENT. An arc is PART OF THE CIRCUMFERENCE.
Assuming any quadrilateral in a circle is cyclic — Only quadrilaterals with ALL FOUR vertices on the circle. Check: opposite angles must sum to 180°.

9. AP SSC Exam Focus

Topic	Marks
Angle at centre vs circumference	3-4
Perpendicular from centre to chord	3-4
Cyclic quadrilateral	4-5
Combined theorem problems	4-5

Proof Strategy for Circle Theorems

Angle at centre = 2× angle at circumference: Draw radius to the point on the circumference. Use isosceles triangle properties (OA=OB=OC=radii). Apply exterior angle theorem.
Angle in semicircle = 90°: This is a special case of Theorem 1 where the angle at centre = 180° (diameter is a straight line).
Cyclic quadrilateral: Join vertices to centre. Use Theorem 1. Sum of opposite angles = ½(360°) = 180°. 'For AP exams: MEMORISE the theorem statements WORD FOR WORD. For proofs: draw the diagram, state the GIVEN, state TO PROVE, then write the proof step by step.'

Worked Example — Cyclic Quadrilateral

In a cyclic quadrilateral ABCD, ∠A = (3x+10)° and ∠C = (2x+30)°. Find all angles. ∠A + ∠C = 180° → (3x+10)+(2x+30) = 180 → 5x+40 = 180 → 5x = 140 → x = 28. ∠A = 3(28)+10 = 94°. ∠C = 2(28)+30 = 86°. Check: 94+86 = 180 ✓. If ∠B = 70°, then ∠D = 110° (opposite angles supplementary).

More Worked Examples

Example — Finding Chord Length: In a circle of radius 13 cm, a chord is at a distance of 5 cm from the centre. Find the length of the chord. Half-chord = √(r² − d²) = √(169−25) = √144 = 12 cm. Full chord length = 24 cm. 'The right triangle formed by radius, perpendicular distance, and half-chord is the KEY to all chord problems. Use Pythagoras.'

Example — Proving a Cyclic Quadrilateral: In quadrilateral ABCD, ∠DAB = 70° and ∠BCD = 110°. Is ABCD cyclic? ∠DAB + ∠BCD = 70° + 110° = 180°. Since a pair of opposite angles sum to 180° → ABCD IS cyclic. 'ONE pair of opposite angles summing to 180° is SUFFICIENT to prove a quadrilateral is cyclic.'

Example — Angle in Same Segment: In a circle, chord AB subtends ∠ACB = 40° on the major arc. P is another point on the same major arc. Find ∠APB. Angles in the same segment are equal → ∠APB = ∠ACB = 40°.

Example — Combined Theorem Problem: In a circle, AB is a diameter. C is a point on the circle such that ∠CAB = 30°. Find ∠CBA. Since AB is diameter → ∠ACB = 90° (angle in semicircle). In ΔABC: ∠CAB + ∠CBA + ∠ACB = 180° → 30° + ∠CBA + 90° = 180° → ∠CBA = 60°.

Complete Circle Theorem Summary

#	Theorem	Key Application
1	Angle at centre = 2 × angle at circumference	Finding unknown angles
2	Angle in a semicircle = 90°	Proving right angles
3	Perpendicular from centre bisects chord	Finding chord length (Pythagoras)
4	Equal chords equidistant from centre	Comparing chords
5	Opposite angles of cyclic quadrilateral = 180°	Finding unknown angles, proving cyclic
6	Angles in same segment are equal	Proving angles equal

Theorem Proof — Angle at Centre (Full Proof for AP Exam)

Given: Circle with centre O. Arc AB subtends ∠AOB at centre and ∠APB at point P on the remaining circle. To Prove: ∠AOB = 2∠APB. Construction: Join PO and extend it to Q (on the circle). Proof: In ΔAOP, OA = OP (radii) → ΔAOP is isosceles → ∠OAP = ∠OPA. Exterior angle ∠AOQ = ∠OAP + ∠OPA = 2∠OPA ... (1). In ΔBOP, OB = OP → ΔBOP is isosceles → ∠OBP = ∠OPB. Exterior angle ∠BOQ = ∠OBP + ∠OPB = 2∠OPB ... (2). Adding (1) and (2): ∠AOQ + ∠BOQ = 2(∠OPA + ∠OPB) → ∠AOB = 2∠APB. 'This proof uses the EXTERIOR ANGLE property of triangles. The AP exam values this proof — practice drawing the three cases (centre on same side, centre inside, centre outside the angle).'

Key Exam Tips

For chord problems: ALWAYS draw the perpendicular from centre to chord. The right triangle gives you the Pythagoras relation: r² = d² + (c/2)².
For cyclic quadrilateral: The test is opposite angles sum to 180°. ONE pair is enough.
'Angle in the same segment' applies when angles are subtended by the SAME chord and lie on the SAME side of the chord.
When you see a diameter in the problem, immediately write '∠ = 90°' at the point on the circle.

Quick Self-Test

In a circle, angle at centre = 100°. Angle at circumference by same arc? (Answer: 50°.)
AB is a diameter. ∠CAB = 25°. Find ∠ABC. (Answer: 65° — angle in semicircle = 90°.)
Opposite angles of a cyclic quadrilateral? (Answer: Sum to 180°.)
Chord = 24 cm, perpendicular distance from centre = 5 cm. Radius? (Answer: 13 cm — use 5²+12²=r².)
Quadrilateral ABCD has ∠A+∠C = 180°. Is it cyclic? (Answer: Yes.)

Key formulas & results

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

Circle Theorems

DEFINITIONS: Circle = all points at fixed distance (radius r) from centre O. Chord = any line segment joining two points on circle. Diameter = longest chord (2r, through centre). Arc = part of circumference. Sector = region between two radii + arc. Segment = region between chord + arc. THEOREM 1 — ANGLE AT CENTRE: ∠AOB (at centre) = 2 × ∠APB (at circumference), where A, B on circle, P on remaining arc. THEOREM 2 — SEMICIRCLE: If AB is a diameter, ∠APB = 90° for any P on circle. (Special case of T1: centre angle = 180°, so circumference angle = 90°.) THEOREM 3 — PERP FROM CENTRE TO CHORD: If OM ⊥ AB (M on chord AB), then AM = MB (chord bisected). Converse: line from centre to midpoint of chord ⊥ chord. USE: right triangle OMB → OB² = OM² + MB². So r = √(d² + (c/2)²) where d = perpendicular distance, c = chord length. THEOREM 4 — EQUAL CHORDS: Equal chords are equidistant from centre. Chords equidistant from centre are equal. THEOREM 5 — CYCLIC QUADRILATERAL: If ABCD is cyclic (all 4 vertices on circle): ∠A + ∠C = 180°. ∠B + ∠D = 180°. Converse: if opposite angles sum to 180°, quadrilateral is cyclic. THEOREM 6 — SAME SEGMENT: Angles in the SAME segment (same side of chord) are equal. WORKED EXAMPLE 1 (T1): ∠AOB = 80° → ∠APB = 40°. WORKED EXAMPLE 2 (T3): chord = 16 cm, dist = 6 cm → r² = 6² + 8² = 100 → r = 10 cm. WORKED EXAMPLE 3 (T5): ∠A = (2x+10)°, ∠C = (3x−20)°, ∠A+∠C=180° → x=38. ∠A=86°, ∠C=94°.

AP EXAM KEY TRAPS: (1) Angle IN A SEMICIRCLE = 90° (at circumference). NOT 180°. 180° is the angle at the CENTRE (the diameter subtends a straight angle at centre, which halves to 90° at circumference). (2) For T3 (perp from centre): HALF the chord for Pythagoras — use MB not AB. (3) In cyclic quadrilateral: OPPOSITE angles sum to 180°. Adjacent angles do NOT necessarily sum to 180°. (4) Theorems apply to arcs on the SAME circle. (5) For proofs: draw the figure first, mark given angles, join centre to endpoints, use isosceles triangle property (OA=OB=OC=r).

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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

✗ Saying the angle in a semicircle is 180°

✓ The angle SUBTENDED BY THE DIAMETER at the CENTRE of the circle is 180° (because the diameter is a straight line). However, the question is usually about the angle SUBTENDED AT ANY POINT ON THE CIRCUMFERENCE by the diameter — and THIS angle is 90° (a right angle). Proof: By the angle at centre theorem, the angle at circumference = ½ × angle at centre = ½ × 180° = 90°. So: if AB is a diameter and P is ANY point on the circle (not on the diameter), then ∠APB = 90°. This is a VERY useful theorem: if you know a triangle is inscribed in a semicircle (one side is the diameter), the angle opposite the diameter is ALWAYS 90°. This can be used to find unknown angles in such triangles without other information.

Practice problems

Work through this chapter's problems as a readiness check — reveal each solution, mark yourself honestly, and get your gap report at the end.

Readiness check

Are you exam-ready for Circles?

1 problems from this chapter. Try each one, reveal the worked solution, mark yourself honestly — get your gap report at the end.

1 questions~2 min

5-minute revision

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

•CIRCLE DEFINITIONS: CIRCLE = set of all points at fixed distance (radius r) from a centre O. CHORD = line segment joining any two points on the circle. DIAMETER = longest chord (passes through centre, length 2r). ARC = part of circumference. SECTOR = region bounded by two radii and an arc. SEGMENT = region bounded by a chord and an arc.
•THEOREM 1 — ANGLE AT CENTRE: The angle subtended by an arc at the CENTRE is TWICE the angle subtended at any point on the remaining circumference. ∠AOB (at centre) = 2 × ∠APB (at circumference). This is the most fundamental circle theorem.
•THEOREM 2 — ANGLE IN SEMICIRCLE: If AB is a diameter and P is ANY point on the circle (other than A or B), then ∠APB = 90°. Special case of Theorem 1: angle at centre = 180° (straight line) → angle at circumference = 90°.
•THEOREM 3 — PERPENDICULAR FROM CENTRE TO CHORD: The perpendicular from the centre of a circle to a chord BISECTS the chord. If OM ⊥ AB where M is on AB, then AM = MB. CONVERSE: the line joining the centre to the midpoint of a chord is perpendicular to the chord.
•USING T3 WITH PYTHAGORAS: In right triangle OMB (right angle at M): OB² = OM² + MB². So r² = d² + (c/2)² where r = radius, d = perpendicular distance from centre to chord, c = full chord length. Solving: r = √[d² + (c/2)²].
•THEOREM 4 — EQUAL CHORDS: Equal chords of a circle are equidistant from the centre. Conversely, chords equidistant from the centre are equal in length. So chord length depends only on distance from centre.
•THEOREM 5 — CYCLIC QUADRILATERAL: If a quadrilateral ABCD has all four vertices on a circle (cyclic), then opposite angles are SUPPLEMENTARY: ∠A + ∠C = 180°. ∠B + ∠D = 180°. CONVERSE: if opposite angles of a quadrilateral sum to 180°, the quadrilateral is cyclic (can be inscribed in a circle).
•THEOREM 6 — ANGLES IN SAME SEGMENT: Angles subtended by a chord on the SAME SIDE (same segment) of the circle are EQUAL. If P and Q lie on the same arc (cut off by chord AB), then ∠APB = ∠AQB.
•DIAMETER as LONGEST CHORD: among all chords of a circle, the diameter is the LONGEST (length 2r). Any chord that does not pass through the centre is shorter than the diameter.
•CONGRUENT CIRCLES: Two circles are CONGRUENT if and only if they have EQUAL RADII. Arcs of congruent circles are equal if and only if the corresponding chords (or central angles) are equal.

Andhra Pradesh (BIEAP) marks blueprint

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

Where this shows up in the real world

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

Bridge arches and circular structures

Many bridge arches and tunnel openings use CIRCULAR ARCS for structural strength (arches efficiently distribute weight downward). The properties of perpendicular from centre, chord-radius relationships, and inscribed angle theorems are used by civil engineers to design these structures. AP's railway and highway bridges, particularly across Krishna and Godavari rivers, use circular arc geometry from this chapter.

Mechanical engineering — gears and bearings

Mechanical gears are CIRCULAR with teeth around the circumference. Bearings (ball bearings, roller bearings) use the property that the perpendicular from centre bisects chords — to ensure even pressure distribution. Cyclic quadrilateral properties are used in gear ratio calculations. AP's manufacturing industry (Visakhapatnam Steel Plant, automotive component factories) employs engineers who use circle geometry daily.

Astronomy and celestial mechanics

Ancient astronomers (Aryabhata, Brahmagupta) used circular orbits to model planetary motion. Although modern physics (Kepler, Newton) showed orbits are elliptical, circular geometry remains essential for understanding angular size, parallax, and apparent motion. AP's planetariums (Vizag) teach these concepts. Satellite orbits are nearly circular and use the circle theorems studied here.

Exam strategy

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

Theorem statement (1-2 marks each): MEMORISE the 6 theorems word-for-word. When asked to state a theorem, use the EXACT mathematical language: 'The angle subtended by an arc at the centre is twice the angle subtended at any point on the remaining part of the circumference.'
Angle calculations (3-4 marks): identify which theorem applies. If central + circumference angles given, use Theorem 1 (2× rule). If diameter mentioned, use Theorem 2 (semicircle = 90°). State the theorem before applying.
Perpendicular from centre + Pythagoras (3 marks): ALWAYS use HALF the chord length (c/2), not full chord. Set up the right triangle with legs OM (distance) and MB (half chord), hypotenuse OB (radius). Apply Pythagoras explicitly: r² = d² + (c/2)².
Cyclic quadrilateral (3-4 marks): set up the equation ∠A + ∠C = 180°. Substitute the algebraic expressions. Solve for the variable. Compute all four angles. Verify: sum of all four = 360°.
Diagram drawing: ALWAYS draw the circle with given information clearly marked. Label centre O, points on the circle, chord, angle. Examiners look at the diagram first — clear diagrams earn structure marks.

Going beyond the textbook

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

Research POWER OF A POINT theorem — for any point P and any circle, if two chords (or secants) through P intersect the circle at points A,B and C,D respectively, then PA × PB = PC × PD. This extends to tangents: if PT is tangent and PAB is secant, then PT² = PA × PB. This is one of the most useful theorems in olympiad geometry.
Investigate the NINE-POINT CIRCLE — for any triangle, there is a special circle passing through 9 notable points: 3 midpoints of sides, 3 feet of altitudes, and 3 midpoints of segments from each vertex to the orthocentre. The nine-point circle has radius = (1/2) × circumradius. Feuerbach's theorem says it is tangent to the incircle and excircles.
Explore PTOLEMY'S THEOREM (cyclic quadrilateral): for a cyclic quadrilateral ABCD, AC × BD = AB × CD + AD × BC (product of diagonals = sum of products of opposite sides). This was used by ancient Greek astronomers to compute trigonometric tables.
Research INVERSIVE GEOMETRY — a transformation centred at point O with radius k maps every point P to a point P' on ray OP with OP × OP' = k². Under inversion, circles through O map to lines, circles not through O map to other circles. Inversive geometry elegantly proves many classical circle theorems. Research the Apollonius problem (constructing a circle tangent to three given circles).

Where else this chapter is tested

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

AP Board SSC (Class 10) — Tangents and Secants to a CircleVery High — Class 9 circle theorems directly extend to Class 10 tangent properties, alternate segment theorem

JEE Main and Advanced (Coordinate Geometry — Circles)Very High — circle properties, chord-radius relations, and inscribed angles are core JEE topics

NTSE (Mathematics)High — circle theorems are standard NTSE topics

Mathematics Olympiad (RMO, INMO)Very High — circle geometry, cyclic quadrilaterals, and inscribed angle theorems are central to olympiad geometry

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PROOF: Consider a circle with diameter AB and a point P anywhere on the circle (not on AB). The angle ∠APB subtended by the diameter at P is what we want to find. APPLY ANGLE AT CENTRE THEOREM: The chord AB subtends an angle of 180° at the CENTRE O (because AOB is a straight line — the diameter passes through O). The angle at the CIRCUMFERENCE (at point P) on the remaining arc = (1/2) × 180° = 90°. Therefore ∠APB = 90° for ANY P on the circle (except A and B). USE: this gives an immediate right angle in any triangle inscribed in a semicircle — very useful for finding unknown angles or applying Pythagoras theorem. EXAMPLE: a triangle inscribed in a semicircle with one side as the diameter is ALWAYS a right triangle with the 90° angle opposite the diameter.

USE THE PERPENDICULAR FROM CENTRE THEOREM + PYTHAGORAS. Setup: chord AB of length c, distance from centre to chord = d (perpendicular distance OM). The perpendicular from centre BISECTS the chord, so AM = MB = c/2. In right triangle OMB: OM = d (perpendicular distance), MB = c/2 (half chord), OB = r (radius, hypotenuse). Pythagoras: OB² = OM² + MB² → r² = d² + (c/2)². EXAMPLE: chord = 16 cm, distance = 6 cm. r² = 6² + (16/2)² = 36 + 64 = 100. r = 10 cm. ALWAYS use HALF the chord for Pythagoras — not the full chord. This is the most common error in these problems.

CYCLIC QUADRILATERAL: a quadrilateral whose ALL FOUR VERTICES lie on a single circle. Equivalently, it can be INSCRIBED in a circle. Not every quadrilateral is cyclic — only those satisfying special angle conditions. KEY PROPERTY: opposite angles of a cyclic quadrilateral are SUPPLEMENTARY (sum to 180°). ∠A + ∠C = 180° and ∠B + ∠D = 180°. CONVERSE (used to PROVE a quadrilateral is cyclic): if opposite angles sum to 180°, it is cyclic. WHICH QUADRILATERALS ARE CYCLIC? Squares and rectangles always (their vertices are on a circle whose diameter equals the diagonal). Isosceles trapeziums always. Rhombi only if they are squares. General parallelograms only if they are rectangles. Cyclic quadrilaterals appear in physics (capacitor design), engineering (gear teeth), and astronomy (planetary positions).

Use the CONVERSE of the semicircle theorem: if a right triangle is inscribed in a circle, the hypotenuse is a diameter. PROOF SKETCH: in right triangle ABC with right angle at B, the circumscribed circle has its centre at the midpoint M of AC (the hypotenuse). Why? Because in a right triangle, the median to the hypotenuse is half the hypotenuse, so BM = AM = CM = (1/2)AC. All three vertices are at distance (1/2)AC from M → all three on a circle with centre M and radius AC/2. The diameter is AC. So the right angle at B is the angle subtended by the diameter AC at point B — exactly the semicircle theorem. USE: this gives an elegant way to construct the circumscribed circle of a right triangle (just bisect the hypotenuse).

Consider chord AB. Points P, Q, R on the same arc (same segment) — all on the same side of chord AB. The angles ∠APB, ∠AQB, ∠ARB are all subtended by chord AB. PROOF: by the Angle at Centre theorem, each of these angles equals (1/2) × ∠AOB (the central angle subtended by AB). Since the central angle is fixed, all the circumference angles in the same segment must be equal. INTUITION: as you move P around the arc (staying on the same side), the angle stays constant — you can change position but the angle 'seen' is the same. This is the basis of the inscribed angle theorem in many proofs. EXAMPLE: if ∠APB = 40° and Q is on the same arc as P, then ∠AQB = 40° (without computing). Saves work in multi-point problems.

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