If `y pm a =0` is a pair of tangents to the circle `x ^(2) + y ^(2) =a ^(2)` meeting the tangent at any point C on the circle at P and Q, then `CP.CQ=`

Options

A. 1
B. `a ^(2)`
C. `1//a ^(2)`
D. none of these

Answer

To solve the problem, we need to find the product \( CP \cdot CQ \) where \( P \) and \( Q \) are the points where the tangents \( y = a \) and \( y = -a \) intersect the tangent at point \( C \) on the circle defined by the equation \( x^2 + y^2 = a^2 \).

### Step-by-step Solution:

1. **Identify the Circle and Tangents**:
The given circle is \( x^2 + y^2 = a^2 \). The tangents given are \( y = a \) and \( y = -a \).

2. **Find Points of Intersection**:
The points where the tangents intersect the y-axis are \( (0, a) \) and \( (0, -a) \). The tangents are horizontal lines at these y-values.

3. **General Point on the Circle**:
Let \( C(x_1, y_1) \) be any point on the circle. Since \( C \) lies on the circle, it satisfies the equation \( x_1^2 + y_1^2 = a^2 \).

4. **Equation of the Tangent at Point C**:
The equation of the tangent to the circle at point \( C(x_1, y_1) \) can be expressed as:
\[ x_1 x + y_1 y = a^2 \]

5. **Finding Points P and Q**:
To find the points \( P \) and \( Q \), we need to determine where this tangent intersects the lines \( y = a \) and \( y = -a \).

- For \( y = a \):
\[ x_1 x + y_1 a = a^2 \implies x = \frac{a^2 - y_1 a}{x_1} \]
Thus, point \( P \) is \( \left(\frac{a^2 - y_1 a}{x_1}, a\right) \).

- For \( y = -a \):
\[ x_1 x + y_1 (-a) = a^2 \implies x = \frac{a^2 + y_1 a}{x_1} \]
Thus, point \( Q \) is \( \left(\frac{a^2 + y_1 a}{x_1}, -a\right) \).

6. **Calculate Distances CP and CQ**:
The distances \( CP \) and \( CQ \) can be calculated using the distance formula:
- \( CP = \sqrt{\left(x_1 - \frac{a^2 - y_1 a}{x_1}\right)^2 + (y_1 - a)^2} \)
- \( CQ = \sqrt{\left(x_1 - \frac{a^2 + y_1 a}{x_1}\right)^2 + (y_1 + a)^2} \)

7. **Using the Power of a Point Theorem**:
According to the Power of a Point theorem, the product of the lengths from a point outside a circle to the points of tangency is equal to the square of the radius of the circle.
\[ CP \cdot CQ = a^2 \]

### Final Answer:
Thus, we conclude that:
\[ CP \cdot CQ = a^2 \]