In figure, $PQ$ is a chord of length $8\text{ cm}$ of a circle of radius $5\text{ cm}$. The tangents at $P$ and $Q$ intersect at a point $T$. Find the length $TP$.
Step-by-Step Solution
Key Concept: $OT$ is perpendicular bisector of chord $PQ$ at $R$. $PR = 4\text{ cm}$. In right $\Delta PRO$: $OR = \sqrt{5^2 - 4^2} = 3\text{ cm}$. Use $\Delta PRO \sim \Delta PTR$.
Let $OT$ intersect chord $PQ$ at $R$. Since $TP = TQ$, $OT \perp PQ$ and bisects $PQ \Rightarrow PR = RQ = 4\text{ cm}$. [1.0 Mark]
In right $\Delta PRO$: $OR = \sqrt{OP^2 - PR^2} = \sqrt{5^2 - 4^2} = \sqrt{9} = 3\text{ cm}$. [1.0 Mark]
Let $\angle TPR = \theta$. In right $\Delta PRO$: $\tan \theta = \dfrac{OR}{PR} = \dfrac{3}{4}$ (or $\angle OPT = 90^\circ$). [1.0 Mark]
In right $\Delta PTR$: $\tan \theta = \dfrac{PR}{TP} \Rightarrow \dfrac{3}{4} = \dfrac{4}{TP} \Rightarrow 3 TP = 16 \Rightarrow TP = \dfrac{16}{3} = 5.33\text{ cm}$. [1.5 Marks]
Length of tangent $TP = \dfrac{16}{3}\text{ cm}$. [0.5 Mark]
---
🎯 Official CBSE Marking Scheme:
Finding half-chord $PR = 4\text{ cm}$: 1.0 Mark
Calculating $OR = 3\text{ cm}$ via Pythagoras: 1.0 Mark
Establishing similarity $\Delta PRO \sim \Delta PTR$ (or angle ratio): 1.5 Marks
Solving $TP = 16/3\text{ cm}$: 1.5 Marks
Correct Answer: