P is a point outside a circle with centre O, and it is 14 cm away from the centre. A secant PAB drawn from P intersects the circle at the points A and B such that PA = 10 cm and PB = 16 cm. The diameter of the Circle is:

Options

A. 10 cm
B. 13 cm
C. 12 cm
D. 11 cm

Answer

To solve the problem, we need to find the diameter of the circle using the given information about the secant and the distances involved. Here's the step-by-step solution:

### Step 1: Understand the Given Information
- Point P is outside the circle, and the distance from P to the center O of the circle is 14 cm.
- The secant PAB intersects the circle at points A and B.
- The lengths are given as PA = 10 cm and PB = 16 cm.

### Step 2: Use the Secant-Tangent Theorem
According to the secant-tangent theorem (or the power of a point theorem), we have:
\[ PA \times PB = PO^2 - r^2 \]
Where:
- \( PA = 10 \) cm
- \( PB = 16 \) cm
- \( PO = 14 \) cm (distance from point P to the center O)
- \( r \) is the radius of the circle.

### Step 3: Calculate PA × PB
First, calculate the product of PA and PB:
\[ PA \times PB = 10 \times 16 = 160 \]

### Step 4: Set Up the Equation
Now, we can set up the equation using the theorem:
\[ 160 = PO^2 - r^2 \]
Substituting \( PO = 14 \):
\[ 160 = 14^2 - r^2 \]
\[ 160 = 196 - r^2 \]

### Step 5: Rearrange the Equation
Rearranging the equation to solve for \( r^2 \):
\[ r^2 = 196 - 160 \]
\[ r^2 = 36 \]

### Step 6: Solve for the Radius
Taking the square root of both sides:
\[ r = \sqrt{36} = 6 \text{ cm} \]

### Step 7: Calculate the Diameter
The diameter \( D \) of the circle is twice the radius:
\[ D = 2r = 2 \times 6 = 12 \text{ cm} \]

### Final Answer
The diameter of the circle is **12 cm**.

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