`PQ` is a chord of length 8 cm, of a circle with centre O and of radius 5 cm. The tangents at P and Q intersect at a point T. The length of TP is

Options

A. `20/3 cm`
B. `21/4 cm`
C. `10/3 cm`
D. `15/4 cm`

Answer

To find the length of \( TP \) in the given problem, we can follow these steps:

### Step 1: Understand the Geometry
We have a circle with center \( O \) and radius \( 5 \) cm. The chord \( PQ \) has a length of \( 8 \) cm. We need to find the length of the tangent \( TP \) where \( T \) is the point where the tangents at points \( P \) and \( Q \) intersect.

### Step 2: Find the Midpoint of Chord \( PQ \)
Let \( R \) be the midpoint of chord \( PQ \). Since \( PQ = 8 \) cm, we have:
\[ PR = RQ = \frac{PQ}{2} = \frac{8}{2} = 4 \text{ cm} \]

### Step 3: Use the Right Triangle
In triangle \( PRO \), where \( O \) is the center of the circle, we can apply the Pythagorean theorem. Since \( OP \) is the radius, we have:
\[ OP = 5 \text{ cm} \quad \text{(radius)} \]
\[ PR = 4 \text{ cm} \quad \text{(half of the chord)} \]
We need to find \( OR \):
\[ OR^2 + PR^2 = OP^2 \]
Substituting the known values:
\[ OR^2 + 4^2 = 5^2 \]
\[ OR^2 + 16 = 25 \]
\[ OR^2 = 25 - 16 = 9 \]
\[ OR = \sqrt{9} = 3 \text{ cm} \]

### Step 4: Apply Similar Triangles
Now, consider triangles \( TPR \) and \( PRO \). Since \( TP \) is tangent to the circle at point \( P \), angle \( TPR \) is \( 90^\circ \). Thus, triangles \( TPR \) and \( PRO \) are similar by AA (Angle-Angle) similarity.

### Step 5: Set Up the Proportions
From the similarity of triangles:
\[ \frac{TP}{PO} = \frac{PR}{OR} \]
Substituting the known lengths:
\[ \frac{TP}{5} = \frac{4}{3} \]

### Step 6: Solve for \( TP \)
Cross-multiplying gives:
\[ TP \cdot 3 = 4 \cdot 5 \]
\[ 3TP = 20 \]
\[ TP = \frac{20}{3} \text{ cm} \]

### Final Answer
Thus, the length of \( TP \) is \( \frac{20}{3} \) cm.

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