NCERT · Class 10 Mathematics · Chapter 10

Circles Full concept notes, theorem by theorem — then the tuition paper's Q3–Q8 solved in full.

Source: jemh110 · Reprint 2026-27 2 theorems 6 solved paper questions Total paper marks covered: 17

Part 1 — Concepts

1. A circle and a line: the only three possibilities

Take a circle and any straight line in the same plane. There are exactly three things that can happen:

No common point — non-intersecting

Two common points — secant

One common point — tangent

Non-intersecting line: no common point with the circle.
Secant: cuts the circle at two distinct points A and B (the part inside is a chord).
Tangent: touches the circle at exactly one point. That point is the point of contact.

Why only three? A line can meet a circle in at most two points — so 0, 1 or 2 is the complete list.

2. Tangent to a circle

A tangent is a line that intersects the circle at only one point. Two facts give you the whole picture:

A tangent is the limiting case of a secant. Fix a line through a point P on the circle and rotate it. The second intersection point slides along the circle, closer and closer to P. At the instant the two endpoints of the chord coincide, you have a tangent. Rotate further and it becomes a secant again.
At any one point of a circle there is exactly one tangent. (Two tangents parallel to a given secant exist — one on each side — but at a single point, only one.)

Physical picture: a wheel rolling on a road. The road is a tangent to the wheel, and the radius drawn to the point where the wheel touches the ground looks exactly perpendicular to the road. That observation is Theorem 10.1.

Theorem 10.1

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

OQ is longer than OP for every Q ≠ P on the tangent

Proof

Given: circle with centre O, tangent XY touching at P. To prove: OP ⊥ XY.
Take any point Q on XY with Q ≠ P, and join OQ.
Q must lie outside the circle — if it were inside, XY would cut the circle twice and be a secant, not a tangent.
So OQ > OP (a point outside the circle is farther from the centre than the radius).
This holds for every point Q on XY except P, so OP is the shortest distance from O to the line XY.
The shortest segment from a point to a line is the perpendicular. Therefore OP ⊥ XY. ∎

Two consequences you will use constantly

∠OPT = 90° at every point of contact P — that single right angle is the engine behind almost every question in this chapter.
Pythagoras in the tangent triangle: since ∠OPT = 90°, OT² = OP² + PT², i.e. PT² = OT² − r². If OT = d and PT = length of tangent ℓ, then ℓ = √(d² − r²).
The line through the point of contact along the radius is also called the normal to the circle at that point.

3. How many tangents from a point?

The answer depends only on where the point sits relative to the circle.

Point P is…	Number of tangents through P	Why
Inside the circle	0	Every line through P meets the circle twice.
On the circle	Exactly 1	Theorem 10.1 — one tangent at a point.
Outside the circle	Exactly 2	Two symmetric touch points, one on each side of OP.

Length of a tangent from an external point P = the length of the segment PT between P and the point of contact T. Notice this is a length of a segment, not of the infinite line.

Theorem 10.2

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

PQ = PR from the external point P

Proof

Given: circle with centre O, external point P, tangents PQ and PR with Q, R the points of contact. To prove: PQ = PR.
Join OP, OQ, OR.
∠OQP = 90° and ∠ORP = 90° (Theorem 10.1 — radius ⊥ tangent).
In right triangles OQP and ORP: OQ = OR (radii of the same circle), OP = OP (common), and both have a right angle.
∴ ∆OQP ≅ ∆ORP by RHS congruence.
∴ PQ = PR (CPCT). ∎

Alternative (Pythagoras): PQ² = OP² − OQ² = OP² − OR² = PR², so PQ = PR.

What else falls out of Theorem 10.2 — these are your weapons

OP bisects the angle between the tangents. Because ∆OQP ≅ ∆ORP, ∠OPQ = ∠OPR and ∠QOP = ∠ROP.
OP is the perpendicular bisector of the chord of contact QR. TP = TQ and OP = OQ ⇒ OT is the perpendicular bisector of PQ (used in Q4 below).
∠QPR + ∠QOR = 180°. In quadrilateral OQPR two opposite angles are 90°, and the angles of a quadrilateral sum to 360°, so the other two must add to 180°. The angle between the two tangents is supplementary to the angle at the centre. If a question gives you one, you instantly have the other.
Identical tangent lengths from a vertex — for any polygon circumscribing a circle, the two tangent segments from each vertex are equal. That is the whole proof of Q6, Q7 and Q8.

4. The three worked examples in the chapter

These three are the actual models for the tuition paper — two of them are paper questions. Read the reasoning pattern, not just the numbers.

Example 1 · Concentric circles (paper Q2)

Two concentric circles C₁ (bigger) and C₂ (smaller) with centre O. A chord AB of C₁ touches C₂ at P. Prove AP = BP.

AB is a tangent to C₂ at P, OP is its radius ⇒ OP ⊥ AB (Theorem 10.1).
AB is a chord of C₁, and the perpendicular from the centre to a chord bisects it ⇒ AP = BP. ∎

Example 2 · ∠PTQ = 2∠OPQ (paper Q3)

Tangents TP, TQ from external point T. Prove ∠PTQ = 2∠OPQ.

Let ∠PTQ = θ. By Theorem 10.2, TP = TQ ⇒ ∆TPQ is isosceles.
∠TPQ = ∠TQP = ½(180° − θ) = 90° − θ/2.
∠OPT = 90° (Theorem 10.1).
∠OPQ = ∠OPT − ∠TPQ = 90° − (90° − θ/2) = θ/2.
∴ ∠PTQ = θ = 2∠OPQ. ∎

Example 3 · Chord 8 cm, radius 5 cm (paper Q4)

PQ is a chord of length 8 cm of a circle of radius 5 cm. Tangents at P and Q meet at T. Find TP.

Join OT; it is the perpendicular bisector of PQ ⇒ PR = RQ = 4 cm.
OR = √(OP² − PR²) = √(25 − 16) = 3 cm.
Right ∆TRP ~ right ∆PRO (AA), so TP/PO = RP/RO ⇒ TP/5 = 4/3 ⇒ TP = 20/3 cm.
Alternative: set TP = x, TR = y; solve x² = y² + 16 and x² + 25 = (y + 3)² ⇒ y = 16/3, x = 20/3.

The one thing that trips everyone up: the "perpendicular from the centre bisects the chord" theorem is from Class 9 — it is allowed, and Exam 1 above leans on it. Keep it in your toolkit alongside 10.1 and 10.2.

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Part 2 — Tuition paper solutions (Q3 – Q8)

Paper: Circles (NCERT) — Unit Test, 45 min. Q1 and Q2 excluded as instructed. Q1 and Q2 are Example-style proofs in Part 1 above anyway (Q2 = Example 1, verbatim).

Q33 marks

Two tangents TP and TQ are drawn to a circle with centre O from an external point T. Prove that ∠PTQ = 2∠OPQ.

Construction: join OP, OQ, PQ. Let ∠PTQ = θ.

TP = TQ — lengths of tangents from an external point (Theorem 10.2). So ∆TPQ is isosceles with TP = TQ.
Base angles of an isosceles triangle are equal: ∠TPQ = ∠TQP.
Angle sum of ∆TPQ: θ + ∠TPQ + ∠TQP = 180° ⇒ ∠TPQ = ∠TQP = ½(180° − θ) = 90° − θ/2.
OP ⊥ TP at the point of contact (Theorem 10.1) ⇒ ∠OPT = 90°.
∠OPQ = ∠OPT − ∠TPQ = 90° − (90° − θ/2) = θ/2.
Therefore ∠PTQ = θ = 2 × (θ/2) = 2∠OPQ. ∎

Answer

∠PTQ = 2∠OPQ, proved. (Equivalently ∠OPQ = ½∠PTQ — the radius into a touch point bisects the angle the tangent makes with the chord of contact.)

Q43 marks

PQ is a chord of length 8 cm of a circle of radius 5 cm. The tangents at P and Q intersect at T. Find TP and TQ.

O–R–T are collinear; OT ⊥ PQ

Given: PQ = 8 cm, OP = OQ = 5 cm. Join OT meeting PQ at R.
TP = TQ (Theorem 10.2) and OP = OQ, so OT is the perpendicular bisector of PQ ⇒ ∠ORT = 90° and PR = RQ = 4 cm.
In right ∆ORP: OR = √(OP² − PR²) = √(25 − 16) = 3 cm.
In right ∆PRT: ∠TPR + ∠PTR = 90°. Also ∠TPR + ∠RPO = ∠TPO = 90° (Theorem 10.1). Hence ∠PTR = ∠RPO.
So ∆TRP ~ ∆PRO by AA (∠TRP = ∠PRO = 90°, ∠PTR = ∠RPO).
Corresponding sides: TP/PO = RP/RO ⇒ TP/5 = 4/3 ⇒ TP = 20/3 cm.
TQ = TP = 20/3 cm (equal tangents).

Answer

TP = TQ = 20/3 cm ≈ 6.67 cm. Check: RT = 16/3, OT = 3 + 16/3 = 25/3, and OT² = 625/9 = 25 + 400/9 = OP² + TP² ✓

Q53 marks

XY and X′Y′ are two parallel tangents to a circle with centre O, and another tangent AB with point of contact C intersects XY at A and X′Y′ at B. Prove that ∠AOB = 90°.

OA and OB bisect the angles at A and B

Construction: let XY touch the circle at P and X′Y′ at Q. Join OP, OQ, OC, OA, OB.

OP ⊥ XY, OQ ⊥ X′Y′ and OC ⊥ AB (Theorem 10.1). So ∠OPA = ∠OCA = 90°.
∆OAP ≅ ∆OAC by RHS: OP = OC (radii), OA common, both right-angled.
∴ ∠OAP = ∠OAC — that is, OA bisects ∠PAB.
Similarly ∆OBQ ≅ ∆OBC (OQ = OC, OB common, right angles) ⇒ OB bisects ∠QBA.
XY ∥ X′Y′ with transversal AB ⇒ ∠PAB + ∠QBA = 180° (interior angles on the same side of a transversal).
Substituting: 2∠OAB + 2∠OBA = 180° ⇒ ∠OAB + ∠OBA = 90°.
In ∆AOB: ∠AOB = 180° − (∠OAB + ∠OBA) = 180° − 90° = 90°. ∎

Answer

∠AOB = 90°, proved. The tangent at the point between two parallel tangents is always seen at a right angle from the centre.

Q62 marks

Prove that a parallelogram circumscribing a circle is a rhombus.

Given: parallelogram ABCD circumscribes a circle, touching AB, BC, CD, DA at P, Q, R, S. To prove: ABCD is a rhombus (all sides equal).
Tangents from the same external point are equal (Theorem 10.2): AP = AS, BP = BQ, CQ = CR, DR = DS.
AB + CD = (AP + PB) + (CR + RD) = (AS + DS) + (BQ + QC) = AD + BC.
In a parallelogram opposite sides are equal: AB = CD and AD = BC.
So 2AB = 2AD ⇒ AB = AD.
Adjacent sides equal ⇒ CD = AB = BC = AD — all four sides equal ⇒ ABCD is a rhombus. ∎

Answer

A parallelogram with an incircle has adjacent sides equal, hence is a rhombus. (General fact to remember: for any quadrilateral circumscribing a circle, AB + CD = AD + BC.)

Q73 marks

Prove that opposite sides of a quadrilateral circumscribing a circle subtend supplementary angles at the centre of the circle.

Given: quadrilateral ABCD circumscribes a circle with centre O, touching AB at P, BC at Q, CD at R, DA at S. Join OA, OB, OC, OD and OP, OQ, OR, OS.
∆OAP ≅ ∆OAS by RHS (OP = OS radii, OA common, ∠OPA = ∠OSA = 90°) ⇒ ∠AOP = ∠AOS.
Similarly: ∠BOP = ∠BOQ (from ∆OBP ≅ ∆OBQ), ∠COQ = ∠COR, ∠DOR = ∠DOS.
Now ∠AOB + ∠COD = (∠AOP + ∠BOP) + (∠COR + ∠DOR).
Replacing each by its equal partner: = ∠AOS + ∠BOQ + ∠COQ + ∠DOS.
And ∠BOC + ∠DOA = (∠BOQ + ∠COQ) + (∠DOS + ∠AOS) — the same four angles.
So the two sums are equal, and together they make up the four angles around O: 2(∠AOB + ∠COD) = 360°.
∴ ∠AOB + ∠COD = 180°, and likewise ∠BOC + ∠DOA = 180°. ∎

Answer

∠AOB + ∠COD = 180° and ∠BOC + ∠DOA = 180° — opposite sides subtend supplementary angles at O.

Q83 marks

A triangle ABC is drawn to circumscribe a circle of radius 4 cm such that the segments BD and DC into which BC is divided by the point of contact D are of lengths 8 cm and 6 cm respectively. Find the sides AB and AC, and the perimeter of ∆ABC.

Let the circle touch AB at F and AC at E. Tangents from a vertex are equal: BF = BD = 8, CE = CD = 6, and let AF = AE = x.
Then AB = x + 8, AC = x + 6, BC = 8 + 6 = 14.
Semi-perimeter s = (x + 8 + x + 6 + 14)/2 = x + 14.
Area = r × s = 4(x + 14).
By Heron's formula: Area² = s(s − a)(s − b)(s − c) = (x + 14)(x)(8)(6) = 48x(x + 14).
Equate: 16(x + 14)² = 48x(x + 14) ⇒ 16(x + 14) = 48x ⇒ x + 14 = 3x ⇒ x = 7.
AB = 7 + 8 = 15 cm, AC = 7 + 6 = 13 cm, BC = 14 cm.
Perimeter = 15 + 13 + 14 = 42 cm.

Answer

AB = 15 cm, AC = 13 cm, BC = 14 cm ⇒ Perimeter = 42 cm. Check: s = 21, Heron area = √(21·7·6·8) = √7056 = 84 = r·s = 4 × 21 ✓

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One-page memory sheet

Theorem 10.1: tangent ⊥ radius at the point of contact → every question starts here.
Theorem 10.2: tangents from an external point are equal → gives isosceles triangles and equal segments everywhere.
Pythagoras: ℓ = √(d² − r²) where d = distance to the external point, ℓ = tangent length.
Angle between tangents + angle at centre = 180°.
OP bisects both the angle between the tangents and the chord of contact.
Quadrilateral circumscribing a circle: AB + CD = AD + BC.
Tangents from one vertex of a circumscribed polygon are equal — label them with one variable and everything solves.
Area shortcut: Area = r × s for any triangle with incircle.