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Class 10 Circles: Core Theorems and Proofs Notes (PDF, MCQ)

Class 10 Circles: Core Theorems and Proofs

This lecture covers the foundational concepts of Class 10 Circles, focusing on tangent properties and related theorems. It provides step-by-step proofs showing that a tangent is perpendicular to the radius at the point of contact and that lengths of tangents from an external point are equal. In addition, multiple board-exam-level problems involving circumscribed triangles, concentric circles, and tangent properties are solved.

Class 10
Maths
English · 7 topics
Checked against the video on 23 Oct 2026

Made by AI from the YouTube video this video (opens YouTube in a new tab), then checked against it. All credit for the teaching goes to the creator. Tap a topic’s time to confirm a fact in the video.

Class 10 Circles: Core Theorems and Proofs

English7 topics · 5 MCQs · 10 flashcardsWatch the video (opens YouTube in a new tab)

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In short

This lecture covers the foundational concepts of Class 10 Circles, focusing on tangent properties and related theorems. It provides step-by-step proofs showing that a tangent is perpendicular to the radius at the point of contact and that lengths of tangents from an external point are equal. In addition, multiple board-exam-level problems involving circumscribed triangles, concentric circles, and tangent properties are solved.

Introduction and Basic Definitions

A circle consists of a center, and connecting the center to any point on the boundary forms a radius.
A line segment joining any two points on a circle is defined as a chord.
Must knowA chord that passes directly through the center of the circle is called the diameter, and its length satisfies d = 2r.
A tangent is a straight line that touches the circle at exactly one point.
The unique point where the tangent line touches the circle is called the point of contact.
Infinitely many tangents can be drawn to a circle across its perimeter.

Theorem: Tangent Perpendicular to Radius

Must knowThe tangent at any point of a circle is perpendicular to the radius through the point of contact.
To prove AB ⊥ OP, take any arbitrary point P' on the tangent AB (other than P) and join it to the center O.
The line segment OP' intersects the circle at Q, meaning OP' = OQ + QP', which implies OP' > OQ.
Since OQ and OP are both radii (OQ = OP = r), it follows directly that OP' > OP.
Because every point on the tangent other than P lies outside the circle, OP is the shortest distance from the center O to the line AB.
The shortest distance from a point to a given line is always the perpendicular distance, proving OP ⊥ AB.
The converse holds true: a line drawn through the endpoint of a radius and perpendicular to it is a tangent to the circle.

Applications of the Perpendicularity Theorem

In a classic problem, tangents drawn at the ends of a diameter of a circle are proved to be parallel to each other.
- Since radius ⊥ tangent, the angles formed at both ends with the diameter are each 90°.
- The consecutive interior angles sum to 90° + 90° = 180° (or alternate interior angles are equal at 90°), proving the tangent lines are parallel.
In another problem with tangent PQ from point P to circle with center O, diameter QOR, and ∠POR = 120°:
- Since QOR is a straight line, ∠POQ = 180° - 120° = 60°.
- Using OQ ⊥ PQ (angle at point of contact = 90°), by angle sum property in △POQ: ∠OPQ = 180° - (90° + 60°) = 30°.

Tangents from an External Point: Theorem & Properties

From an external point lying outside a circle, exactly two tangents can be drawn to the circle.
Must knowThe lengths of the two tangents drawn from an external point to a circle are equal (PQ = PR).
To prove PQ = PR, join the center O to the contact points Q and R, and connect external point P to O.
- In right triangles △PQO and △PRO: OQ = OR (radii), ∠PQO = ∠PRO = 90°, and OP = OP (common hypotenuse).
- △PQO ≅ △PRO by the RHS congruence criterion, yielding PQ = PR by CPCT.
The line joining the external point to the center acts as an angle bisector for both the angle between the tangents (∠QPO = ∠RPO) and the angle subtended at the center (∠QOP = ∠ROP).

Worked Problems on Circumscribed Polygons

In a problem where tangents from P have an angle of 60° between them on a circle of radius a:
- The line OP bisects the 60° angle into two 30° angles.
- In right-angled △PQO, sin(30°) = OQ / OP = a / OP, yielding OP = 2a.
When a triangle ABC circumscribes a circle touching at points, tangents from each vertex are equal in length:
- In △ABC touching BC at Q, AB at P, and AC at R with AP = 4 cm, BP = 3 cm, and AC = 11 cm:
- Tangents from B give BQ = BP = 3 cm; tangents from A give AR = AP = 4 cm.
- Length CR = AC - AR = 11 - 4 = 7 cm, giving QC = CR = 7 cm.
- The length of BC is computed as BQ + QC = 3 + 7 = 10 cm.
When △ABC with AB = AC circumscribes a circle touching BC at E:
- By equal tangents: BD = BE, CF = CE, and AD = AF.
- Substituting into AB = AC (AD + BD = AF + CF) simplifies to BD = CF, which directly proves BE = CE.

Advanced Problems on Tangents and Circles

For a circle touching side BC of △ABC at P and extended sides AB and AC at Q and R:
- Using equal tangent segments from external points A, B, and C gives AQ = AR, BQ = BP, and CR = CP.
- Perimeter of △ABC = AB + BC + CA = AB + (BP + PC) + AC = (AB + BQ) + (AC + CR) = AQ + AR = 2AQ.
- Therefore, AQ = (1/2) × (Perimeter of △ABC).
For an inscribed circle in a right △ABC (∠B = 90°, BC = 48 cm, AB = 14 cm):
- Hypotenuse AC is calculated via Pythagoras theorem: AC = √(14² + 48²) = √2500 = 50 cm.
- Total Area(△ABC) = Area(△OBC) + Area(△OAB) + Area(△OCA).
- (1/2) × 14 × 48 = (1/2) × r × (48 + 14 + 50), which simplifies to 336 = 56r, giving in-radius r = 6 cm.
In a quadrilateral ABCD circumscribing a circle:
- Splitting each side into its tangent segments demonstrates that AB + CD = BC + DA.
Opposite sides of a quadrilateral circumscribing a circle subtend supplementary angles at the center of the circle:
- Joining the center to the vertices and contact points divides the central angles into pairs of equal angles (2∠1 + 2∠2 + 2∠3 + 2∠4 = 360°).
- Hence, (∠1 + ∠4) + (∠2 + ∠3) = 180°, proving ∠AOB + ∠COD = 180°.

Parallel Tangents and Tangent Chord Problems

When two parallel tangents XY and X'Y' are intersected by a third tangent AB with contact point C:
- The sum of consecutive interior angles ∠PAC + ∠QBC = 180°.
- Since the center lines bisect these angles, 2∠1 + 2∠2 = 180°, meaning ∠1 + ∠2 = 90°.
- In △AOB, by the angle sum property: ∠AOB = 180° - (∠1 + ∠2) = 180° - 90° = 90°.
For two tangents TP and TQ drawn from an external point T:
- Let ∠TPQ = ∠TQP = ∠1 (since TP = TQ). In △TPQ, ∠PTQ + 2∠1 = 180°.
- Since radius ⊥ tangent, ∠1 = 90° - ∠OPQ.
- Substituting yields ∠PTQ + 2(90° - ∠OPQ) = 180°, which proves ∠PTQ = 2∠OPQ.
The line joining the external point P to center O is the perpendicular bisector of the chord joining the points of contact.
In concentric circles of radii a and b (where a > b), the length of the chord of the larger circle tangent to the smaller circle is 2√(a² - b²).

Key terms

Tangent: A straight line that touches a circle at exactly one point
Point of Contact: The specific single point where a tangent line meets a circle
Chord: A straight line segment connecting two points on the perimeter of a circle
In-circle: A circle inscribed inside a polygon such that it touches all sides
Circumscribing: Drawing a figure around another so that their boundaries touch without intersecting
Perpendicular Bisector: A line that divides another segment into two equal halves at an angle of 90 degrees

Formulas

d = 2r: Relation between diameter (d) and radius (r) of a circle
sin θ = Perpendicular / Hypotenuse: Trigonometric sine ratio used to find segment lengths in right-angled triangles
tan θ = Perpendicular / Base: Trigonometric tangent ratio used across similar right-angled triangles
Area(△) = (1/2) × Base × Height: Formula used to compute triangle areas when expressing total area as sum of component triangles
Length of chord = 2 × √(a² - b²): Length of the chord of a larger concentric circle tangent to an inner circle of radius b

Exam tips and traps

Whenever a tangent and a radius appear together in a problem, immediately mark the angle between them at the point of contact as 90 degrees.
For problems where a circle is inscribed in a triangle, break the total area into three smaller triangles meeting at the center O to solve for in-radius r.

Quick revision

Read these the night before the exam.

A tangent to a circle is perpendicular to the radius at the point of contact.
Exactly two tangents can be drawn to a circle from a point outside it.
The lengths of tangents drawn from an external point to a circle are always equal.
The line connecting an external point to the circle center bisects the angle between the two tangents.
Tangents drawn at the two opposite ends of a diameter are always parallel.
In a quadrilateral circumscribed around a circle, the sum of opposite sides is equal: AB + CD = BC + DA.
Opposite sides of a quadrilateral circumscribing a circle subtend supplementary angles (summing to 180°) at the center.
The line segment connecting the center to an external point is the perpendicular bisector of the chord of contact.

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