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Circles
RD Sharma
CBSE
Grade 10
Question:
Prove that the lengths of tangents drawn from an external point to a circle are equal (Theorem 10.2).
Step-by-Step Solution
Key Concept: Given: Tangents $TP, TQ$ from $T$. To Prove: $TP = TQ$. Construction: Join $OP, OQ, OT$. Proof: RHS congruence of $\Delta OPT$ and $\Delta OQT$.
Given: A circle $C(O, r)$ and tangents $TP, TQ$ from external point $T$. [0.5 Mark]
Construction: Join $OP, OQ$ and $OT$. [0.5 Mark]
Proof: In right $\Delta OPT$ and right $\Delta OQT$:
$\angle OPT = \angle OQT = 90^\circ$ (Radius $\perp$ Tangent)
$OT = OT$ (Common hypotenuse)
$OP = OQ$ (Radii of same circle). [1.5 Marks]
By RHS congruence criterion, $\Delta OPT \cong \Delta OQT \Rightarrow TP = TQ$ (CPCT). Proved! [0.5 Mark]
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🎯 Official CBSE Marking Scheme:
Given & Construction: 1.0 Mark
Proving RHS congruence of $\Delta OPT$ and $\Delta OQT$: 1.5 Marks
Concluding $TP = TQ$: 0.5 Mark
Construction: Join $OP, OQ$ and $OT$. [0.5 Mark]
Proof: In right $\Delta OPT$ and right $\Delta OQT$:
$\angle OPT = \angle OQT = 90^\circ$ (Radius $\perp$ Tangent)
$OT = OT$ (Common hypotenuse)
$OP = OQ$ (Radii of same circle). [1.5 Marks]
By RHS congruence criterion, $\Delta OPT \cong \Delta OQT \Rightarrow TP = TQ$ (CPCT). Proved! [0.5 Mark]
---
🎯 Official CBSE Marking Scheme:
Given & Construction: 1.0 Mark
Proving RHS congruence of $\Delta OPT$ and $\Delta OQT$: 1.5 Marks
Concluding $TP = TQ$: 0.5 Mark
Correct Answer:
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