If a line is drawn through a point A(3,4) to cut the circle `x^(2)+y^(2)=4` at P and Q then AP .AQ=

Options

A. 15
B. 17
C. 21
D. 25

Answer

To solve the problem, we need to find the product \( AP \cdot AQ \) where \( A(3, 4) \) is a point outside the circle defined by the equation \( x^2 + y^2 = 4 \).

### Step-by-Step Solution:

1. **Identify the Circle and its Radius**:
The equation of the circle is given by:
\[ x^2 + y^2 = 4 \]
This can be rewritten as:
\[ x^2 + y^2 = r^2 \]
where \( r^2 = 4 \). Thus, the radius \( r \) of the circle is:
\[ r = \sqrt{4} = 2 \]

2. **Use the Power of a Point Theorem**:
According to the Power of a Point theorem, for a point \( A \) outside a circle, the product of the lengths of the segments from the point to the points of intersection with the circle is given by:
\[ AP \cdot AQ = OA^2 - r^2 \]
where \( O \) is the center of the circle and \( r \) is the radius.

3. **Calculate the Distance \( OA \)**:
The center \( O \) of the circle is at the origin \( (0, 0) \). The coordinates of point \( A \) are \( (3, 4) \). We can find the distance \( OA \) using the distance formula:
\[ OA = \sqrt{(3 - 0)^2 + (4 - 0)^2} = \sqrt{3^2 + 4^2} = \sqrt{9 + 16} = \sqrt{25} = 5 \]

4. **Substitute Values into the Power of a Point Formula**:
Now we can substitute \( OA \) and \( r \) into the formula:
\[ AP \cdot AQ = OA^2 - r^2 = 5^2 - 2^2 \]
Calculating this gives:
\[ AP \cdot AQ = 25 - 4 = 21 \]

5. **Final Answer**:
Therefore, the value of \( AP \cdot AQ \) is:
\[ \boxed{21} \]