Lesson 6: Circles and Their Properties

Circles generate more board-exam theorems per page than almost any other NCERT chapter. This lesson gathers them in a working order: tangents and their lengths, angles in the alternate segments and semicircles, and cyclic quadrilaterals — each with the proof idea, because CBSE regularly asks for the proof itself.

Tangents from an external point

Three facts, all provable from one diagram:

• The tangent at any point of a circle is perpendicular to the radius through the point of contact.
• The two tangents drawn from an external point are equal in length.
• The line joining the centre to the external point bisects the angle between the tangents.

The workhorse relation is Pythagoras in the triangle (centre, point of contact, external point): (tangent length)² = OP² − r².

Worked example 1 (basic). From a point P at distance 13 cm from the centre O of a circle of radius 5 cm, find the tangent length. Tangent length = √(13² − 5²) = √(169 − 25) = √144 = 12 cm. (Another 5-12-13 triple.)

Worked example 2 (intermediate). Two concentric circles have radii 5 cm and 3 cm. Find the length of the chord of the larger circle that touches the smaller. The chord is tangent to the inner circle, so the inner radius to the contact point is perpendicular to the chord and bisects it. Half the chord = √(5² − 3²) = √16 = 4 cm, so the chord is 8 cm.

Worked example 3 (NCERT exemplar style). Prove that the tangents from an external point P to a circle with centre O are equal. Let the tangents touch at A and B. In △OAP and △OBP: OA = OB (radii), OP is common, and ∠OAP = ∠OBP = 90° (radius ⊥ tangent). So △OAP ≅ △OBP by RHS, giving PA = PB. The same congruence gives ∠APO = ∠BPO — one proof harvests three results, which is exactly why examiners like it.

Angles in circles

• The angle in a semicircle is a right angle: if AB is a diameter, then for any point C on the circle, ∠ACB = 90°.
• Angles in the same segment are equal.
• Cyclic quadrilaterals: opposite angles are supplementary — they sum to 180°.
• The angle between a chord and a tangent equals the angle in the alternate segment.

Angle computation example. In a cyclic quadrilateral ABCD, ∠A = 75° and ∠B = 105°. Then ∠C = 180° − 75° = 105° and ∠D = 180° − 105° = 75°. If instead you are given ∠A : ∠C = 2 : 3, then 2k + 3k = 180° gives k = 36°, so ∠A = 72° and ∠C = 108°.

Competency application: the roundabout

A circular roundabout has radius 20 m. A straight cycle path is built as a chord of the circle, passing 12 m from the centre. How long is the path within the roundabout? Half the chord = √(20² − 12²) = √(400 − 144) = √256 = 16 m, so the path is 32 m. The theorem in use — the perpendicular from the centre to a chord bisects the chord — is the same one as Worked example 2; only the story changed. A genuine competency question is often a theorem wearing a costume; the skill is recognising the costume.

Common mistakes

• Applying Pythagoras with OP as a leg. OP is the *hypotenuse* (it faces the right angle at the contact point). Tangent² = OP² − r², never OP² + r².
• Forgetting the perpendicular from the centre bisects the chord. Students compute the half-chord and report it as the answer; the question asked for the chord — double it.
• Calling any quadrilateral in a circle "cyclic" and using supplementary angles without all four vertices on the circle. The theorem needs all four points concyclic.
• In tangent proofs, citing SSS instead of RHS. The clean NCERT proof uses the right angles; that is what the marking scheme looks for.

Exam Experience

Section A — Multiple choice (1 mark each)

Q1. The length of the tangent from a point at distance 10 cm from the centre of a circle of radius 6 cm is: (a) 4 cm (b) 8 cm (c) 16 cm (d) √136 cm. Answer: (b) 8 cm. √(10² − 6²) = √64 = 8.

Q2. In a cyclic quadrilateral, one angle is 110°. Its opposite angle is: (a) 110° (b) 90° (c) 70° (d) 55°. Answer: (c) 70°. Opposite angles of a cyclic quadrilateral sum to 180°.

Section B — Short answer (3 marks) Q3. Two tangents PA and PB are drawn from an external point P to a circle with centre O. If ∠APB = 60°, find ∠AOB and the length PA given OP = 14 cm and radius 7 cm. Marking scheme: 1 mark for ∠AOB = 180° − 60° = 120° (quadrilateral OAPB has two right angles at A and B); 1 mark for using PA² = OP² − r² = 196 − 49 = 147; 1 mark for PA = √147 = 7√3 ≈ 12.12 cm.

Section C — Long answer (5 marks) Q4. Prove that the angle in a semicircle is a right angle, and use it in the following: AB is a diameter of a circle with centre O and radius 5 cm. Point C on the circle is such that AC = 6 cm. Find BC. Marking scheme: Proof — 1 mark for the correct diagram/construction (join OC); 1 mark for OA = OB = OC (radii) giving two isosceles triangles; 1 mark for angle chasing: if ∠OCA = x and ∠OCB = y, then 2x + 2y = 180° in △ABC, so ∠ACB = x + y = 90°. Application — 1 mark for recognising △ABC is right-angled at C so BC² = AB² − AC² = 100 − 36 = 64; 1 mark for BC = 8 cm.

Going deeper: the angle at the centre theorem

The angle subtended by an arc at the centre is double the angle subtended at any point on the remaining part of the circle. So if a chord AB subtends 70° at the centre O, it subtends 35° at every point of the major arc — which is also why all angles in the same segment are equal (they are all half of the same central angle). The semicircle theorem of the main lesson is the special case where the central angle is 180° (a straight line, the diameter), halving to 90°. One theorem, three corollaries: learn the parent, inherit the children.

Going deeper: equal chords and their distances

Equal chords of a circle are equidistant from the centre, and chords equidistant from the centre are equal. Computationally this is the same half-chord Pythagoras as before: in a circle of radius 10 cm, a chord of length 16 cm has half-length 8, so its distance from the centre is √(100 − 64) = 6 cm. Two chords of the same circle with lengths 16 cm and 12 cm stand at distances 6 cm and √(100 − 36) = 8 cm respectively — the longer chord sits closer, always.

Going deeper: tangents and the supplementary angle

If PA and PB are tangents from P, then ∠APB + ∠AOB = 180°: the quadrilateral OAPB has right angles at A and B (radius ⊥ tangent), and quadrilateral angles sum to 360°. So tangents meeting at 70° face a central angle of 110°. This also gives a quick route to the angle between a tangent and the line OP: it is half of ∠AOB by the isosceles triangle OAP… — more precisely, in △OAP the angles are 90° at A, (∠AOB)/2 at O and (180° − ∠AOB)/2 at P, since OP bisects both angles. These micro-facts are the connective tissue of exemplar circle proofs; collecting them turns three-mark proofs into one-mark observations.

One more family: cyclic quadrilateral drills

Two facts settle most cyclic-quadrilateral questions instantly. Opposite angles are supplementary: if ∠A = 105° then ∠C = 75°. And the exterior angle equals the interior opposite angle: if a side is extended so the exterior angle is 80°, the interior angle at the far vertex is also 80°. Proof of the second in one line: exterior = 180° − adjacent interior, and adjacent interior + opposite interior = 180°, so exterior = opposite interior. Equal arcs also cut equal chords and equal angles at the circumference — three interchangeable currencies (arc, chord, angle) that exemplar problems constantly convert between.

Final drills

Drill 1 — the power of a point. From an external point P, a tangent PT touches a circle and a secant PAB cuts it at A and B. Then PT² = PA × PB. With PT = 6 cm and PA = 4 cm: PB = 36/4 = 9 cm (so the chord AB = PB − PA = 5 cm). This result — a consequence of similar triangles formed by the alternate-segment angle — is the exemplar chapter's favourite "unseen" relation; it also explains why the tangent length formula works: the secant through the centre gives PA × PB = (OP − r)(OP + r) = OP² − r² = PT².

Drill 2 — tangent–chord angle. A tangent at A meets chord AB at 50°. The angle in the alternate segment — the angle in the opposite arc, e.g. ∠ACB for C on the far side — equals 50°. Meanwhile the angle in the *near* segment is supplementary: 130°. Examiners ask for one and mark the other as the distractor; identify which segment is "alternate" (the one on the far side of the chord from the angle) before answering.