In the given 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 of $TP$.
Step-by-Step Solution
Key Concept: Let $OT$ intersect $PQ$ at $R$. $PR = 4\text{ cm}, OR = \sqrt{5^2 - 4^2} = 3\text{ cm}$. Right $\Delta ORP \sim$ right $\Delta OPT \Rightarrow \dfrac{TP}{PO} = \dfrac{RP}{RO} \Rightarrow \dfrac{TP}{5} = \dfrac{4}{3} \Rightarrow TP = \dfrac{20}{3}\text{ cm}$. Alternatively, Pythagoras in $\Delta PRT$: $TP^2 = TR^2 + 16$. In $\Delta OPT$: $TP^2 + 25 = (TR + 3)^2 \Rightarrow TR^2 + 16 + 25 = TR^2 + 6TR + 9 \Rightarrow 6TR = 32 \Rightarrow TR = 16/3 \Rightarrow TP = 20/3\text{ cm}$.
Perpendicular $OR \perp PQ \Rightarrow PR = RQ = 4\text{ cm}$. In $\Delta ORP$, $OR = \sqrt{25 - 16} = 3\text{ cm}$. [1.5 Marks]
Let $TR = y$. In right $\Delta PRT$, $TP^2 = y^2 + 16$. (1) [1.0 Mark]
In right $\Delta OPT$, $TP^2 + 5^2 = (y + 3)^2 \Rightarrow y^2 + 16 + 25 = y^2 + 6y + 9$. [1.0 Mark]
$6y = 32 \Rightarrow y = \dfrac{16}{3}\text{ cm} \Rightarrow TP = \sqrt{\left(\dfrac{16}{3}\right)^2 + 16} = \sqrt{\dfrac{256 + 144}{9}} = \dfrac{20}{3}\text{ cm}$. [1.5 Marks]
---
🎯 Official CBSE Marking Scheme:
Finding $PR = 4\text{ cm}, OR = 3\text{ cm}$: 1.5 Marks
Forming Pythagorean equations in $\Delta PRT$ and $\Delta OPT$: 2.0 Marks
Solving $TR = 16/3 \Rightarrow TP = 20/3\text{ cm}$: 1.5 Marks
Correct Answer: