A line is drawn through the point `P(3,11)` to cut the circle `x^(2)+y^(2)=9` at A and B. Then `PA.PB` is equal to

Options

A. 121
B. 74
C. 65
D. 53

Answer

To solve the problem, we need to find the product of the lengths PA and PB, where P is the point (3, 11) and A and B are the points where a line through P intersects the circle defined by the equation \(x^2 + y^2 = 9\).

### Step-by-Step Solution:

1. **Identify the Circle's Equation**:
The equation of the circle is given as \(x^2 + y^2 = 9\). This represents a circle centered at the origin (0, 0) with a radius of 3 (since \(r^2 = 9\)).

2. **Use the Power of a Point Theorem**:
According to the Power of a Point theorem, if a point P lies outside a circle, the product of the lengths from the point to the points of intersection (A and B) of a line through P and the circle is equal to the square of the length of the tangent from P to the circle. This can be expressed as:
\[ PA \cdot PB = PT^2 \]
where \(PT\) is the length of the tangent from point P to the circle.

3. **Calculate the Length of the Tangent (PT)**:
The formula for the length of the tangent from a point \((x_1, y_1)\) to a circle \(x^2 + y^2 = r^2\) is given by:
\[ PT = \sqrt{x_1^2 + y_1^2 - r^2} \]
Here, \(P(3, 11)\) and the radius \(r = 3\). Thus, we substitute \(x_1 = 3\), \(y_1 = 11\), and \(r^2 = 9\):
\[ PT = \sqrt{3^2 + 11^2 - 9} \]

4. **Calculate \(PT^2\)**:
First, calculate \(3^2\) and \(11^2\):
\[ 3^2 = 9, \quad 11^2 = 121 \]
Now, substitute these values into the equation:
\[ PT = \sqrt{9 + 121 - 9} = \sqrt{121} = 11 \]
Therefore, \(PT^2 = 11^2 = 121\).

5. **Conclusion**:
Since \(PA \cdot PB = PT^2\), we have:
\[ PA \cdot PB = 121 \]

Thus, the final answer is:
\[ \boxed{121} \]