Circles

1. Circle terminology — complete glossary

Before diving into theorems, every NCERT term must be locked in.

• Centre (O): the fixed point from which all points on the circle are equidistant.
• Radius (r): the constant distance from the centre to any point on the circle. Every radius of a given circle is equal.
• Diameter (d = 2r): a chord that passes through the centre. It is the longest chord of the circle.
• Chord: a line segment whose both endpoints lie on the circle. A diameter is a special chord.
• Arc: a continuous piece of the circle cut off by a chord. Two types:
  • Minor arc: the shorter piece — subtends a central angle less than 180 degrees.
  • Major arc: the longer piece — subtends a central angle greater than 180 degrees.
  • Semicircle: exactly half the circle — each arc subtends 180 degrees at the centre.
• Segment: region enclosed between a chord and its arc.
  • Minor segment: the smaller region (between chord and minor arc).
  • Major segment: the larger region (between chord and major arc).
• Sector: the pizza-slice region bounded by two radii and an arc. A semicircle is a special sector.
• Concentric circles: circles with the same centre but different radii.
• Congruent circles: circles with equal radii.
• Concyclic points: points that all lie on a single circle.

2. Perpendicular from centre to chord bisects it (Theorem 9.1)

Statement: The perpendicular from the centre of a circle to a chord bisects the chord.

Given: Circle with centre O. AB is a chord. OM is perpendicular to AB, so angle OMA = angle OMB = 90 degrees.

To prove: AM = MB.

Construction: Join OA and OB.

Proof:

1. In triangles OMA and OMB:
   • OA = OB (radii of the same circle)
   • OM = OM (common side)
   • angle OMA = angle OMB = 90 degrees (given)
2. By the RHS congruence rule: triangle OMA is congruent to triangle OMB.
3. By CPCT: AM = MB. Hence M is the midpoint of AB. Square.

Converse (Theorem 9.2): The line drawn through the centre of a circle to bisect a chord is perpendicular to the chord.

Proof of converse: Given AM = MB. In triangles OMA and OMB: OA = OB (radii), AM = BM (given), OM = OM (common). By SSS congruence, triangle OMA is congruent to triangle OMB. So angle OMA = angle OMB. Since they are supplementary angles on a straight line, each must equal 90 degrees. Hence OM is perpendicular to AB. Square.

3. Equal chords are equidistant from the centre (Theorem 9.3)

Statement: Equal chords of a circle (or of congruent circles) are equidistant from the centre (or centres).

Given: Circle with centre O. AB = CD. OM perpendicular to AB and ON perpendicular to CD.

To prove: OM = ON.

Proof:

1. OM perpendicular to AB means M is the midpoint of AB (Theorem 9.1), so AM = AB/2.
2. ON perpendicular to CD means N is the midpoint of CD, so CN = CD/2.
3. Since AB = CD, we get AM = CN.
4. In triangles OAM and OCN:
   • OA = OC (radii)
   • AM = CN (step 3)
   • angle OMA = angle ONC = 90 degrees
5. By RHS congruence: triangle OAM is congruent to triangle OCN. By CPCT: OM = ON. Square.

Converse (Theorem 9.4): Chords equidistant from the centre of a circle are equal in length.

Proof of converse: Given OM = ON. In triangles OAM and OCN: OA = OC (radii), OM = ON (given), angle OMA = angle ONC = 90 degrees. By RHS, the triangles are congruent. By CPCT: AM = CN, so AB = CD. Square.

4. Equal chords subtend equal angles at the centre (Theorems 9.5 and 9.6)

Theorem 9.5: Equal chords of a circle subtend equal angles at the centre.

Proof: Let AB = CD. In triangles AOB and COD: OA = OC, OB = OD (all radii), AB = CD (given). By SSS congruence, triangle AOB is congruent to triangle COD. By CPCT, angle AOB = angle COD. Square.

Theorem 9.6 (Converse): If two chords of a circle subtend equal angles at the centre, the chords are equal.

Proof: Given angle AOB = angle COD. In triangles AOB and COD: OA = OC, OB = OD (radii), angle AOB = angle COD (given). By SAS congruence, triangle AOB is congruent to triangle COD. By CPCT, AB = CD. Square.

5. Angle at centre = 2 times angle at remaining circle (Theorem 9.8)

This is the most important theorem in the chapter — almost everything else follows from it.

Theorem 9.8: The angle subtended by an arc at the centre is double the angle subtended by it at any point on the remaining part of the circle.

Given: Arc PQ of a circle with centre O. R is any point on the major arc. Prove that angle POQ = 2 times angle PRQ.

Proof (case: O lies inside triangle PRQ):

1. Join R to O and extend the line to point A on the circle.
2. In triangle OPR: OP = OR (radii), so the triangle is isosceles. Hence angle ORP = angle OPR. The exterior angle at O along the line RA gives: angle POR' = angle ORP + angle OPR = 2 times angle ORP. ...(i)
3. Similarly in triangle OQR: angle QOR' = 2 times angle ORQ. ...(ii)
4. Adding (i) and (ii): angle POQ = angle POR' + angle QOR' = 2(angle ORP + angle ORQ) = 2 times angle PRQ. Square.

The same argument works for all three cases: minor arc (O inside the angle), semicircle, and major arc (O outside the angle — subtract instead of add). The result holds in all cases.

6. Angle in a semicircle is 90 degrees

This is the most famous corollary of Theorem 9.8, known as Thales' Theorem.

Corollary: The angle subtended by a diameter at any point on the circle is a right angle (90 degrees).

Proof: Let AB be a diameter with centre O. Let P be any point on the circle (not A or B). AB subtends angle AOB = 180 degrees at O (straight line). By Theorem 9.8: $$\angle APB = \frac{1}{2} \times \angle AOB = \frac{180}{2} = 90^{\circ}.$$ This holds for every position of P on the semicircle. Square.

Converse: If an angle inscribed in a circle is 90 degrees, then the chord forming the angle is a diameter.

7. Angles in the same segment are equal (Theorem 9.9)

Theorem 9.9: Angles in the same segment of a circle are equal.

Given: Minor arc PQ. R and S are two points on the major arc (same segment). Prove angle PRQ = angle PSQ.

Proof:

1. Both angle PRQ and angle PSQ are subtended by the same arc PQ at points on the major arc.
2. By Theorem 9.8: angle POQ = 2 times angle PRQ (R on major arc).
3. Also: angle POQ = 2 times angle PSQ (S on major arc, same central angle).
4. From steps 2 and 3: 2 times angle PRQ = 2 times angle PSQ, so angle PRQ = angle PSQ. Square.

Converse (Theorem 9.10): If a line segment joining two points subtends equal angles at two other points on the same side of the line, then all four points are concyclic.

8. Cyclic quadrilateral — opposite angles sum to 180 degrees (Theorem 9.11)

A cyclic quadrilateral is a quadrilateral all of whose four vertices lie on a circle. The circle is its circumcircle.

Theorem 9.11: The sum of either pair of opposite angles of a cyclic quadrilateral is 180 degrees.

Given: Cyclic quadrilateral ABCD inscribed in a circle with centre O.

To prove: angle DAB + angle BCD = 180 degrees and angle ABC + angle CDA = 180 degrees.

Proof:

1. Arc BCD subtends angle BOD at the centre and angle BAD at point A (on arc BAD). By Theorem 9.8: angle BOD = 2 times angle BAD. ...(i)
2. Arc BAD subtends the reflex angle BOD at the centre and angle BCD at point C (on arc BCD). By Theorem 9.8: reflex angle BOD = 2 times angle BCD. ...(ii)
3. Adding (i) and (ii): angle BOD + reflex angle BOD = 2 times angle BAD + 2 times angle BCD.
4. Left side = 360 degrees (complete angle at O). So 2(angle BAD + angle BCD) = 360 degrees.
5. Therefore angle DAB + angle BCD = 180 degrees. Square.
6. Since all four angles sum to 360 degrees: angle ABC + angle CDA = 360 - 180 = 180 degrees. Square.

9. Converse — when is a quadrilateral cyclic? (Theorem 9.12)

Theorem 9.12: If the sum of a pair of opposite angles of a quadrilateral is 180 degrees, then the quadrilateral is cyclic.

Proof sketch: Let quadrilateral ABCD have angle A + angle C = 180 degrees. Draw the circle through A, B, D. Suppose C does not lie on this circle; let C' be the point on arc BD (other than A) such that ABDC' is cyclic. By Theorem 9.11, angle A + angle C' = 180 degrees. But angle A + angle C = 180 degrees as well, so angle C' = angle C. This forces C' = C, meaning C lies on the circle. Hence ABCD is cyclic. Square.

Key consequences for exam:

• A rectangle is always cyclic (each angle is 90 degrees; opposite angles sum to 180 degrees).
• A square is always cyclic.
• A parallelogram is cyclic only if it is a rectangle.
• An isosceles trapezium is always cyclic.

10. All theorems at a glance

11. Common mistakes to avoid

• Using the full chord length instead of half the chord when applying Pythagoras — the perpendicular from the centre bisects the chord, so use AM (half), not AB (full).
• Confusing which arc is subtending the angle at the centre — always identify the arc first, then decide if you need the ordinary or reflex central angle.
• Assuming all quadrilaterals are cyclic — only those where opposite angles sum to 180 degrees qualify. A general parallelogram is NOT cyclic.
• Forgetting to write CPCT explicitly in congruence proofs — CBSE expects this step to be named.
• Mixing up segment (a region) and arc (a curve) — "angle in a segment" means the vertex is on the arc forming the boundary of that segment.
• In the angle-at-centre theorem, forgetting that the inscribed angle must be on the remaining arc (not the same arc).