PYQs
Practise questions grouped by chapter and source year.
Practice questions are organised using the years and labels in our source collection. Work through the solutions with your teacher; these are not a complete official examination paper.
Practice question 1
Source year: 20253 marksDraw tangents to circle radius 5cm from point 13cm away. [CBSE 2025]
Practice question 2
Source year: 20243 marksProve tangent ⊥ radius. [CBSE 2024]
Practice question 3
Source year: 20253 marksDraw tangents to circle of radius 5cm from point 13cm away from center. Measure length of tangents. [CBSE 2025]
Practice question 4
Source year: 20243 marksProve that tangent at any point of circle is perpendicular to radius through point of contact. [CBSE 2024]
Practice question 5
Source year: 20233 marksFrom external point, two tangents are drawn to circle. Prove they are equal in length. [CBSE 2023]
Practice question 6
Source year: 20222 marksA tangent PQ at point P of circle of radius 5cm meets line through center O at Q such that OQ = 13cm. Find PQ. [CBSE 2022]
Formulae
Revisit important relationships and quick revision formulae.
Tangent Length
PT = √(d²-r²)
Tangent Properties
- Tangent ⊥ radius at point of contact\n2. From external point, two tangents are equal\n3. Length of tangent = √(d²-r²) where d=distance from center, r=radius
Secant Properties
Line intersecting circle at two points\nChord: Part of secant inside circle
Derivations
Work through the reasoning behind key concepts.
Tangent Length Formula
From external point T to circle with center O, radius r. At point of contact P: OP ⊥ PT (tangent perpendicular to radius). In △OPT: OT² = OP² + PT² (Pythagoras). So PT² = OT² - OP² = d² - r². Therefore PT = √(d²-r²).
Tangent Length Formula
In △OPT (O=center, P=point of contact, T=external point):\nOP ⊥ PT (tangent perpendicular to radius)\nBy Pythagoras: OT² = OP² + PT²\nPT² = OT² - OP² = d² - r²\nPT = √(d²-r²)
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