Q1. [1] § 10.4 Summary
In the given figure, $QR$ is a common tangent to the two given circles touching externally at $A$. The tangent at $A$ meets $QR$ at $P$. If $AP = 4.2$ cm, then the length of $QR$ is :
- A 4.2 cm
- B 2.1 cm
- C 8.4 cm
- D 6.3 cm
Previously asked in CBSE board exam
2024 30/4/1 Q18
Generated by claude-sonnet-4-6 · 2026-06-15 10:23 · grounding rag
Model Answer
(C) 8.4 cm
By the equal tangent theorem, from external point P: PQ = PA = 4.2 cm and PR = PA = 4.2 cm. Therefore, QR = PQ + PR = 4.2 + 4.2 = 8.4 cm.
Source: Chapter 10, Theorem 10.2
Explanation
P is an external point to both circles. Tangents from P to the first circle: PA = PQ (equal tangents). Tangents from P to the second circle: PA = PR (equal tangents). So QR = QP + PR = 2 × 4.2 = 8.4 cm. Examiners expect you to cite the equal-tangents theorem.
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