Q1. [3] § 10.2 Tangent to a Circle § 10.3 Number of Tangents from a Point on a Circle § 10.4 Summary
Draw a circle of radius 3 cm. From a point P lying outside the circle at a distance of 6 cm from its centre, construct two tangents PA and PB to the circle.
Previously asked in CBSE board exam
2022 30/3/1 Q7(b) (OR-2)
Generated by claude-sonnet-4-6 · 2026-06-15 10:25 · grounding rag
Model Answer
Steps of Construction:
- Draw a circle with centre O and radius 3 cm.
- Mark a point P such that OP = 6 cm.
- Find the midpoint M of OP (by bisecting OP).
- Draw a circle with centre M and radius MO (= MP = 3 cm).
- This circle intersects the original circle at points A and B.
- Join PA and PB. These are the required tangents.
Justification: ∠OAP = 90° (angle in a semicircle), so OA ⊥ PA. Since OA is a radius and PA ⊥ OA, PA is a tangent. Similarly, PB is a tangent.
$$PA = PB = \sqrt{OP^2 - OA^2} = \sqrt{36 - 9} = \sqrt{27} = 3\sqrt{3} \text{ cm}$$
Source: Chapter 10, Sections 10.2 and 10.3
---
Explanation
- Key idea: A tangent is perpendicular to the radius at the point of contact, so ∠OAP = 90°. This means A lies on a circle with OP as diameter (angle in a semicircle = 90°).
- Step to remember: Always bisect OP to find centre M, then draw the auxiliary circle of radius OP/2.
- Length verification using Pythagoras is expected and earns a mark.
- Examiners award marks for: correct steps (1 mark), accurate diagram (1 mark), justification/measurement (1 mark).
If a question refers to an image, map, graph or diagram that is not shown here, find the actual CBSE question paper on the CBSE website:
cbse.gov.in.