NCERT Class 10 Maths Chapter 10 Circles Formula Sheet 2026-27

The Class 10 Maths Chapter 10 Circles formula sheet collects every theorem, result and formula on a single page. Circles rests on just two theorems. The tangent is perpendicular to the radius at the point of contact. The two tangents from an external point are equal in length. Every board question follows from these two facts plus Pythagoras and basic congruence. The formulas match the 2026-27 CBSE syllabus.

• Two governing theorems: tangent perpendicular to radius (Theorem 10.1) and equal tangents from an external point (Theorem 10.2).
• Core formula: length of tangent PT = √(OP² − r²), derived from Pythagoras on the right triangle at the point of contact.
• CBSE board applications: circumscribed quadrilateral (opposite sides equal), angle between tangents plus central angle equals 180°, and the concentric-circle chord bisection result.

Watch Circles Class 10 Maths Explained

Source: Magnet Brains on YouTube

All Circles Formulas: Complete List

The table below lists every formula, theorem and result you need. It mirrors the single-page PDF. Two theorems do almost all the work: the right angle at the point of contact (Theorem 10.1) and the equal-tangents result (Theorem 10.2). Every board question uses one or both, plus Pythagoras or congruence.

Tangent Perpendicular to Radius (Theorem 10.1)

The first governing theorem says the tangent at any point of a circle is perpendicular to the radius at that point of contact: ∠OPT = 90°. This single right angle turns every tangent problem into a right-triangle problem where Pythagoras applies at once (PT² + r² = OP²). The NCERT proof: of all segments from centre O to the tangent line, the radius OP is the shortest, and the shortest distance from a point to a line is the perpendicular.

The practical step: whenever a tangent appears in a diagram, immediately draw the radius to the point of contact and mark the right angle. That one action turns the geometry into a right triangle. Forgetting it is the most common setup error in board answers.

Length of Tangent from an External Point

The most used formula is PT = √(OP² − r²). Radius r and tangent PT are the legs of a right triangle and OP is the hypotenuse, so Pythagoras gives PT at once.

A useful check: the tangent length is always shorter than OP (PT < OP), since PT is a leg and OP the hypotenuse. If your answer gives PT > OP, there is an arithmetic error.

Equal Tangents from an External Point (Theorem 10.2)

The second governing theorem powers the polygon problems. The two tangents from any external point to a circle are equal: PQ = PR. NCERT proves it by RHS congruence: in right triangles OQP and ORP, OQ = OR (radii) and OP is the common hypotenuse, so they are congruent and PQ = PR by CPCT. Two bonus results follow: OP bisects ∠QPR, and ∠QPR + ∠QOR = 180°.

The key application is a circle inscribed in a polygon. Label the equal tangent segments from each vertex with the same letter. For a triangle, if tangents from A, B, C have lengths a, b, c, then AB = a + b, BC = b + c, CA = c + a. This equal-labelling trick solves every circumscribed polygon problem in the chapter.

Angle Between Two Tangents

When two tangents are drawn from an external point P, the angle between them (∠QPR) and the central angle (∠QOR) are supplementary: ∠QPR + ∠QOR = 180°. Proof: in quadrilateral OQPR the angles at Q and R are both 90°, so the other two sum to 180°. This is the fast formula for all "find the angle" MCQs.

So if the central angle is 130°, then ∠QPR = 180° − 130° = 50°. Students who memorise this solve such 2-mark questions in seconds. The bisector relation also helps: OP bisects ∠QPR, so ∠OPQ = ∠QPR / 2.

Circumscribed Quadrilateral & Triangle

These Exercise 10.2 results account for the highest-value board questions. They all follow from equal tangents (Theorem 10.2) by adding up equal segments.

For the triangle incircle problem, let BD = BF = x, CD = CE = y, AE = AF = z. Then AB = x + z, BC = x + y, CA = y + z. Solve these three equations to find any unknown segment.

Concentric Circles: Chord Bisection Formula

For two concentric circles (same centre O, radii R and r with R > r), a chord of the outer circle that is a tangent to the inner circle is bisected at the point where it touches the inner circle.

Why it is bisected: OT (inner radius) is perpendicular to chord AB by Theorem 10.1, and a perpendicular from the centre bisects any chord. So AT = TB = √(R² − r²), and the full chord is twice this. The common mistake is forgetting to double the half-chord when the question asks for the full chord.

CBSE Board Exam Weightage

Circles is a high-value chapter that tests both theorem proofs and numerical work. The table below shows the question types that appear.

Chapter 10 on its own usually accounts for a 3-mark or 4-mark question. The circumscribed quadrilateral and the tangent-length calculation are the two most predictable high-value items.

Common Mistakes to Avoid

More Class 10 Circles Resources

Use this formula sheet alongside the other Chapter 10 Circles resources below. Each covers a different aspect of the chapter, so together they give complete board preparation.

All Class 10 Maths Formula Sheets