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MathematicsCh-10 15 min comprehensive revision
NCERT Class 10 Mathematics — Chapter 10

Circles

Tangents to a circle, Theorem 10.1 (radius perpendicular to tangent at point of contact), Theorem 10.2 (lengths of tangents from external point are equal), and circumscribed polygon riders.

Quick Key Takeaways:
Definition of Tangent: A straight line that touches the circle at exactly one point (point of contact). A circle can have infinitely many tangents, but only two parallel tangents at the endpoints of a diameter.
Theorem 10.1 (Radius-Tangent Perpendicularity): The tangent at any point of a circle is perpendicular to the radius through the point of contact: OP⊥XYOP \perp XY.
Theorem 10.2 (Lengths of External Tangents): The lengths of tangents drawn from an external point to a circle are equal: PA=PBPA = PB.
Circumscribing Quadrilateral Property: If a quadrilateral ABCDABCD circumscribes a circle, then: AB+CD=AD+BCAB + CD = AD + BC
Jump to Concept:1. 1. Circle Geometry, Tangent Properties & Theorem 10.12. 2. Theorem 10.2 Proof & High-Yield Geometric Riders3. 3. Important Solved Board Examination Questions (3-Mark & 5-Mark)4. 4. CBSE Case-Study Modeling: Triangle Circumscribing a Circle (Radius 4 cm)5. 5. CBSE Examiner Marking Scheme, Calculation Traps & Step Rubric
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1. Circle Geometry, Tangent Properties & Theorem 10.1

Axioms & Foundational Theory

Comprehensive mathematical foundations, definitions, formulas, and visual conceptual model for Circles.

• Theorem 10.1: Tangent is Perpendicular to the Radius
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Statement: The tangent at any point of a circle is perpendicular to the radius through the point of contact.
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Given: A circle with centre OO and a tangent XYXY touching the circle at point PP.
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To Prove: OP⊥XYOP \perp XY.
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Proof:
1. Take any point QQ on XYXY other than PP. Join OQOQ.
2. The point QQ must lie outside the circle (if QQ were inside, XYXY would become a secant and intersect the circle at two points).
3. Therefore, OQ>radius OPOQ > \text{radius } OP.
4. Since this is true for every point on line XYXY except PP, OPOP is the shortest distance from centre OO to line XYXY.
5. The shortest distance between a point and a line is the perpendicular distance.
6. Hence, OP⊥XY\mathbf{OP \perp XY}.
📊 Circles: Tangent Theorems & Circumscribed PolygonsVisual Model
OPABKEY THEOREMS1. PA = PBTangents from ext.point are equal.2. OA ⊥ PARadius ⊥ Tangentat contact point.

Visual schematic mapping radius-tangent perpendicularity, external point tangent equality, and circumscribed quadrilateral properties.

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2. Theorem 10.2 Proof & High-Yield Geometric Riders

Theorem Proofs & Derivations

Step-by-step rigorous geometric and algebraic theorem proofs and formula derivations for Circles.

• Theorem 10.2: Tangents from an External Point are Equal
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Statement: The lengths of tangents drawn from an external point to a circle are equal.
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Given: A circle with centre OO, an external point PP, and two tangents PAPA and PBPB touching the circle at AA and BB.
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To Prove: PA=PBPA = PB.
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Construction: Join OP,OAOP, OA, and OBOB.
•
Proof:
1. In right triangles ΔOAP\Delta OAP and ΔOBP\Delta OBP:
- ∠OAP=∠OBP=90∘\angle OAP = \angle OBP = 90^\circ [By Theorem 10.1: radius ⊥\perp tangent]
- OP=OPOP = OP [Common hypotenuse]
- OA=OBOA = OB [Radii of the same circle]
2. By RHS Congruence Criterion, ΔOAP≅ΔOBP\mathbf{\Delta OAP \cong \Delta OBP}.
3. Therefore, by CPCT (Corresponding Parts of Congruent Triangles):
PA=PB\mathbf{PA = PB}
∠OPA=∠OPB(OP is angle bisector)\angle OPA = \angle OPB \quad (OP\text{ is angle bisector})
∠AOP=∠BOP\angle AOP = \angle BOP
• Hence Proved.
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3. Important Solved Board Examination Questions (3-Mark & 5-Mark)

Topper Step Solutions

Standard CBSE board exam numericals and riders with complete step-by-step mathematical working and justifications.

• 3-Mark Standard Board Question: Prove that in two concentric circles, the chord of the larger circle which touches the smaller circle is bisected at the point of contact.
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Step 1 (Given & To Prove):
Two concentric circles C1C_1 (larger) and C2C_2 (smaller) with common centre OO.
ABAB is a chord of C1C_1 touching C2C_2 at point PP.
To prove: AP=PBAP = PB.
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Step 2 (Proof):
1. Since ABAB is tangent to smaller circle C2C_2 at PP and OPOP is the radius: OP⊥ABOP \perp AB [By Theorem 10.1].
2. Now, ABAB is a chord of the larger circle C1C_1 and OP⊥ABOP \perp AB.
3. We know that the perpendicular drawn from the centre of a circle to a chord bisects the chord.
4. Therefore, AP=PB\mathbf{AP = PB}.
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Conclusion: The chord is bisected at the point of contact.
• 5-Mark Heavyweight Board Problem / Rider: A quadrilateral ABCDABCD is drawn to circumscribe a circle. Prove that AB+CD=AD+BCAB + CD = AD + BC.
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Step 1 (Given & Construction):
Let quadrilateral ABCDABCD touch the circle at points P,Q,R,SP, Q, R, S on sides AB,BC,CD,DAAB, BC, CD, DA respectively.
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Step 2 (Applying Theorem 10.2): Lengths of tangents from an external point are equal:
- From vertex AA: AP=AS— (1)AP = AS \quad \text{--- (1)}
- From vertex BB: BP=BQ— (2)BP = BQ \quad \text{--- (2)}
- From vertex CC: CR=CQ— (3)CR = CQ \quad \text{--- (3)}
- From vertex DD: DR=DS— (4)DR = DS \quad \text{--- (4)}
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Step 3 (Adding Equations 1 to 4):
(AP+BP)+(CR+DR)=(AS+DS)+(BQ+CQ)(AP + BP) + (CR + DR) = (AS + DS) + (BQ + CQ)
AB+CD=AD+BCAB + CD = AD + BC
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Conclusion: AB+CD=AD+BC\mathbf{AB + CD = AD + BC} Hence Proved.
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4. CBSE Case-Study Modeling: Triangle Circumscribing a Circle (Radius 4 cm)

Case Study Mastery (4 Marks)

Real-world mathematical modeling scenario with structured multi-part questions and full step solutions.

• Practical Application Context: Triangle Circumscribing a Circle (Radius 4 cm)
A triangle ABCABC is drawn to circumscribe a circle of radius 4 cm such that the segments BDBD and DCDC into which BCBC is divided by the point of contact DD are of lengths 8 cm and 6 cm respectively. Find the sides ABAB and ACAC.
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Q1: Express the lengths of ABAB and ACAC in terms of unknown tangent xx. →\rightarrow Let tangents from AA be AF=AE=xAF = AE = x. Tangents from BB: BD=BF=8BD = BF = 8 cm. Tangents from CC: CD=CE=6CD = CE = 6 cm. Therefore, a=BC=8+6=14a = BC = 8 + 6 = 14 cm, b=AC=x+6b = AC = x + 6 cm, c=AB=x+8c = AB = x + 8 cm.
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Q2: Find the semi-perimeter ss and area using Heron's formula. →\rightarrow s=14+(x+6)+(x+8)2=x+14s = \frac{14 + (x+6) + (x+8)}{2} = x + 14. Area(ΔABC)=s(s−a)(s−b)(s−c)=(x+14)(x)(8)(6)=48x(x+14)\text{Area}(\Delta ABC) = \sqrt{s(s-a)(s-b)(s-c)} = \sqrt{(x+14)(x)(8)(6)} = \mathbf{\sqrt{48x(x+14)}}.
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Q3: Equate with sum of areas of ΔOBC+ΔOCA+ΔOAB\Delta OBC + \Delta OCA + \Delta OAB to find x,AB,ACx, AB, AC. →\rightarrow Area=12r(a+b+c)=r⋅s=4(x+14)\text{Area} = \frac{1}{2} r (a + b + c) = r \cdot s = 4(x + 14). Equating: 48x(x+14)=4(x+14)  ⟹  48x(x+14)=16(x+14)2  ⟹  3x=x+14  ⟹  2x=14  ⟹  x=7 cm\sqrt{48x(x+14)} = 4(x+14) \implies 48x(x+14) = 16(x+14)^2 \implies 3x = x + 14 \implies 2x = 14 \implies \mathbf{x = 7\text{ cm}}. Thus, side AB=x+8=7+8=15 cmAB = x + 8 = 7 + 8 = \mathbf{15\text{ cm}}, and side AC=x+6=7+6=13 cmAC = x + 6 = 7 + 6 = \mathbf{13\text{ cm}}.
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5. CBSE Examiner Marking Scheme, Calculation Traps & Step Rubric

Important Solved Board Questions

Examiner step-marking allocations, common algebraic/arithmetic deduction traps, and presentation guidelines.

• Step-by-Step Marking Rubric & Presentation Guidelines
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1 Mark: Stating Theorem 10.2 (PA=PBPA = PB) with justification.
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1 Mark: Correct RHS congruence proof in Theorem 10.2.
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2 Marks: Systematic algebraic addition of 4 tangent pairs for circumscribed quadrilaterals.
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1 Mark: Final boxed values with units (cm).
• Common Calculation Traps & Verification Checklist
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Trap 1: Forgetting to align tangent pairs on the correct sides of equations (ensure AP+BPAP + BP stay on LHS to form ABAB).
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Trap 2: In RHS congruence, forgetting to state that hypotenuse OPOP is common.
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Trap 3: Assuming a tangent passes through the circle (a secant passes through; a tangent touches at exactly one point).
Authentic Board Question (3 Marks)Topic: Circles Mathematical Proof & Computations
State the governing mathematical theorem/formula, show complete geometric or computational steps, and find the exact result for Circles.

Official CBSE Step-by-Step Marking Breakdown:

Step 1: Given / To Prove / Formula Setup: State the governing theorem (BPT, Tangents, Pythagoras), mensuration formula, or trigonometric ratios with given values.
1 Mark
Step 2: Step-by-Step Derivation / Arithmetic Working: Show complete deductive steps or numerical calculations with π=22/7\pi = 22/7 or trigonometric standard values.
1 Mark
Step 3: Final Boxed Answer with Units / Q.E.D.: State the final numerical result with proper square/cubic units (cm2,m3\text{cm}^2, \text{m}^3) or formal proof conclusion.
1 Mark
Model Student Answer (Target: Full 3/3 Marks):
To score full 3 marks on Circles in CBSE Mathematics:

1. Theorem / Formula: Write the standard governing formula or state the geometric theorem clearly.
2. Calculation / Proof: Substitute dimensions systematically or provide deductive step justifications.
3. Final Result: Box the final answer with required units (cm,cm2,m3\text{cm}, \text{cm}^2, \text{m}^3) or conclude with "Hence Proved".
Examiner Mark Deduction Traps:
•Always write intermediate calculation lines to earn partial credit under CBSE step-marking.
•State reasons in parentheses (e.g., "[Tangents from an external point are equal]") during geometric proofs.

High-Frequency Conceptual Doubts & FAQs

Curated answers to the most common questions asked by Class 10 students.
Theorem 10.1 states: The tangent at any point of a circle is perpendicular to the radius through the point of contact (OP⊥ABOP \perp AB). It establishes a 90∘90^\circ angle at the point of contact, enabling the use of Pythagoras theorem in right ΔOPT\Delta OPT: OT2=OP2+PT2OT^2 = OP^2 + PT^2.

Related YouTube Videos & Masterclasses

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Curated top-tier CBSE Class 10 video lessons, one-shots, and problem-solving sessions for Circles. Click any video below to watch instantly inside Master10.

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Circles One Shot 🔥 | Class 10 Maths Chapter 10 | Ritik Mishra

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