CBSE Class 10 Maths — Circles

Chapter Overview & Weightage

Circles is one of the most formula-dense chapters in CBSE Class 10 Maths, and it regularly appears in the board exam for 6-8 marks — typically one 3-mark question and one 5-mark question, or two 3-mark problems. The chapter builds on Class 9 circle properties and introduces tangents from an external point.

In CBSE 2023 and 2024 board exams, Circles contributed 6-7 marks. The most frequent question type: “prove that tangents from an external point are equal in length” (5 marks) and “find angle in a given tangent-secant configuration” (3 marks). These two alone can fetch 5-6 marks if prepared well.

Year	Marks	Question Type
2024	7	Proof + angle calculation
2023	6	Tangent length + tangent-secant
2022	6	Proof + circle in quadrilateral
2021	8	Extended proof + two problems

Key Concepts You Must Know

Tangent: A line that touches a circle at exactly one point (the point of tangency). At the point of tangency, the tangent is perpendicular to the radius.

Secant: A line that intersects a circle at two points.

Number of tangents:

From a point inside the circle: 0 tangents
From a point on the circle: exactly 1 tangent
From a point outside the circle: exactly 2 tangents

Key theorem: The tangent at any point of a circle is perpendicular to the radius through the point of tangency.

Lengths of tangents: The two tangents drawn from an external point to a circle are equal in length.

Angle between tangent and chord: Equals the inscribed angle in the alternate segment (tangent-chord angle = angle in alternate segment).

Important Formulas

If $P$ is an external point and $O$ is the centre of a circle with radius $r$ , and $PT$ is the tangent from $P$ to point $T$ on the circle:

PT = \sqrt{OP^2 - r^2} = \sqrt{OP^2 - OT^2}

This follows from the Pythagorean theorem since $\angle OTP = 90°$ .

If two tangents PA and PB are drawn from point P:

\angle APB + \angle AOB = 180°

(The quadrilateral AOBP has $\angle OAP = \angle OBP = 90°$ , so the other two angles sum to 180°.)

Angle between tangent PT and chord TQ = Inscribed angle TRQ in the alternate segment.

If from external point P, a tangent PT and a secant PAB are drawn:

PT^2 = PA \times PB

where A and B are the two points where the secant intersects the circle.

Solved Previous Year Questions

PYQ 1 — Proof (5 marks, CBSE 2023)

Q: Prove that the tangents drawn from an external point to a circle are equal in length.

Solution:

Let O be the centre of a circle with radius $r$ . Let P be an external point. Let PA and PB be the two tangents (A and B are points of tangency).

Given: OA = OB = $r$ (radii), $\angle OAP = \angle OBP = 90°$ (radius ⊥ tangent)

To prove: PA = PB

In triangles OAP and OBP:

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

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

Therefore PA = PB (corresponding parts of congruent triangles — CPCT). ✓

Bonus result from the proof: $\angle OPA = \angle OPB$ , so OP bisects $\angle APB$ . Also, OP ⊥ AB (the line joining the external point to the centre is perpendicular to the chord AB).

PYQ 2 — Angle Calculation (3 marks, CBSE 2022)

Q: If PA and PB are tangents from external point P to a circle with centre O, and $\angle APB = 70°$ , find $\angle AOB$ .

Solution:

In quadrilateral AOBP: $\angle OAP + \angle AOB + \angle OBP + \angle APB = 360°$

$90° + \angle AOB + 90° + 70° = 360°$

$\angle AOB = 360° - 250° = 110°$

PYQ 3 — Circle Inscribed in Quadrilateral

Q: Prove that the opposite sides of a quadrilateral circumscribing a circle are supplementary.

Solution (outline):

Let a circle touch the four sides AB, BC, CD, DA at P, Q, R, S respectively.

Since tangents from an external point are equal:

From A: AP = AS
From B: BP = BQ
From C: CQ = CR
From D: DR = DS

Now: AB + CD = (AP + PB) + (CR + RD) = (AS + BQ) + (CQ + DS) = (AS + DS) + (BQ + CQ) = AD + BC

So AB + CD = AD + BC, meaning opposite sides are equal in sum. ✓

Difficulty Distribution

Difficulty	Marks	What’s Tested
Easy (1-2 marks)	~2	Define tangent, number of tangents, basic angle
Medium (3 marks)	~3	Tangent length calculation, angle in tangent problems
Hard (5 marks)	~3	Proofs (tangent equality, circle in quadrilateral)

Expert Strategy

Learn the two big proofs cold. The “tangents from external point are equal” proof and the “opposite sides of circumscribed quadrilateral are supplementary” proof appear in CBSE year after year. A clean, well-labelled proof with proper “Given, To Prove, Construction, Proof” structure gets full 5 marks. Memorise these exactly.

For angle problems, always draw the figure first. Mark all right angles (radius ⊥ tangent). Then look for the angle sum properties: angles in a quadrilateral = 360°, angles in a triangle = 180°. The answer usually follows in 2-3 steps.

For tangent length problems: the setup is always a right triangle OTP (where O = centre, T = point of tangency, P = external point). Apply Pythagoras: $PT = \sqrt{OP^2 - OT^2}$ . This appears in 3-mark questions as a direct calculation.

Combination problems: CBSE sometimes gives a figure with both a tangent and a secant from the same external point. Use $PT^2 = PA \times PB$ . These appear in Case Study questions too.

Common Traps

Trap 1: Writing $\angle OTP = 90°$ only when asked to prove tangent properties. But this right angle must be assumed known when solving angle problems — don’t reprove it every time. In a 3-mark angle calculation, simply state “OT ⊥ PT (radius ⊥ tangent)” and move on.

Trap 2: Confusing the angle between two tangents with the angle subtended at the centre. $\angle APB + \angle AOB = 180°$ only when AOBP is a quadrilateral (two tangents from one external point). If the problem configuration is different, recheck.

Trap 3: In “circle inscribed in quadrilateral” problems, assuming all sides are equal. That’s only true for a rhombus. For a general quadrilateral with an inscribed circle, only the opposite side sums are equal: $AB + CD = BC + DA$ .

Trap 4: Using $PT^2 = PA \times PB$ without checking that P, A, B are collinear (i.e., it’s actually a secant through P). The formula requires A and B to be on the same secant from P.