Class 10 — Circles

Chapter Overview & Weightage

“Circles” in Class 10 deals with tangent properties — a small but high-yield chapter. Every CBSE board paper has at least one question worth 3–5 marks.

CBSE Class 10 Maths — Circles Weightage

Year	Marks	Question Type
2024	5	3-mark proof + 2-mark numerical
2023	4	4-mark long answer
2022	5	1-mark + 4-mark theorem application
2021	3	3-mark single problem
2020	5	5-mark theorem proof

About 6–7% of the maths paper. Reliable scoring.

Key Concepts You Must Know

Ranked by board frequency:

Tangent at any point of a circle is perpendicular to the radius at the point of contact (Theorem 10.1).
Lengths of two tangents from an external point are equal (Theorem 10.2).
Number of tangents from an external point = 2.
Chord properties (carried over from Class 9 for proof construction).
Tangent-secant configuration for circle problems.

Important Formulas

Length of tangent from external point at distance $d$ from centre, circle radius $r$ :

\ell = \sqrt{d^2 - r^2}

Two tangents from external point P make equal lengths PA = PB.

Angle between two tangents from external point + angle subtended by chord AB at centre = 180°.

The only numerical formula is the Pythagorean tangent length. Most problems are pure-geometry proofs.

Solved Previous Year Questions

PYQ 1 — Equal tangents proof (CBSE 2024, 3 marks)

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

Solution. Let O be the centre of the circle, P an external point, and PA, PB the two tangents (A, B on the circle).

Join OA, OB, OP. Since tangents are perpendicular to radii: $\angle OAP = \angle OBP = 90°$ .

In triangles OAP and OBP:

$OA = OB$ (radii)
$OP = OP$ (common)
$\angle OAP = \angle OBP = 90°$

By RHS congruence: $\triangle OAP \cong \triangle OBP$ .

Therefore $PA = PB$ (CPCT). Hence proved.

PYQ 2 — Tangent length numerical (CBSE 2023, 2 marks)

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

Solution.

\ell = \sqrt{d^2 - r^2} = \sqrt{169 - 25} = \sqrt{144} = 12 \text{ cm}

PYQ 3 — Quadrilateral with inscribed circle (CBSE 2022, 4 marks)

Q. Prove that the sum of opposite sides of a quadrilateral circumscribing a circle is equal.

Solution. Let ABCD be the quadrilateral with sides AB, BC, CD, DA touching the circle at P, Q, R, S respectively.

By the equal-tangents theorem from each vertex:

AP = AS (from A)
BP = BQ (from B)
CR = CQ (from C)
DR = DS (from D)

Adding the relevant sums: $AB + CD = AP + PB + CR + RD = AS + BQ + CQ + SD = AD + BC$

Hence $AB + CD = BC + AD$ . Hence proved.

Difficulty Distribution

Easy (30%): Direct application of tangent length formula, “find the length of tangent” type.
Medium (50%): Proofs of standard theorems, two-tangent configuration problems.
Hard (20%): Combined with quadrilaterals or triangles, multi-step constructions.

Expert Strategy

Toppers’ approach:

Memorise the two main theorems word-for-word. Tangent ⊥ radius. Two tangents from external point are equal.
Draw before you solve. Even simple problems need a clean diagram with tangent points labelled.
Recognise the “RHS congruence” pattern. Whenever you have radius–tangent–common-side setup, RHS gives congruence in 2 lines.
Quadrilateral circumscribing a circle is a guaranteed 4-mark question every 2 years. Practise it twice.

Common Traps

Trap 1 — Forgetting to mention CPCT. When proving sides equal via congruence, students show the triangles are congruent but forget to write “CPCT” (corresponding parts of congruent triangles). Half-mark deducted.

Trap 2 — Using wrong congruence rule. Tangent-radius gives 90° angle → use RHS, not SAS. RHS is specifically for right-angled triangles.

Trap 3 — Drawing tangents that look like secants. A tangent touches the circle at exactly one point; a secant cuts at two. Diagram errors cost marks.

Trap 4 — Confusing chord and tangent. Chord is inside the circle. Tangent is outside, touching at one point. The “tangent perpendicular to radius” theorem applies only to tangents.