The tangents PA and PB are drawn from any point P of the circle `x^(2)+y^(2)=2a^(2)` to the circle `x^(2)+y^(2)=a^(2)`. The chord of contact AB on extending meets again the first circle at the points A' and B'. The locus of the point of intersection of tangents at A' and B' may be given as

Options

A. `x^(2)+y^(2)=8a^(2)`
B. `x^(2)+y^(2)=4a^(2)`
C. `x^(2)+y^(2)=6a^(2)`
D. none of these

Explanation

To solve the given problem, we need to find the locus of the point of intersection of the tangents at points A' and B' on the larger circle, given the conditions of the problem. Let's break down the solution step by step.

### Step 1: Identify the circles and their equations
We have two circles:
1. Smaller circle: \( C_1: x^2 + y^2 = a^2 \) (radius = a)
2. Larger circle: \( C_2: x^2 + y^2 = 2a^2 \) (radius = \( \sqrt{2}a \))

### Step 2: Parametrize the point P on the larger circle
Let \( P \) be any point on the larger circle \( C_2 \). We can express the coordinates of point \( P \) using parametric equations:
\[ P = (x_1, y_1) = (\sqrt{2}a \cos \theta, \sqrt{2}a \sin \theta) \]

### Step 3: Find the equation of the chord of contact from point P to the smaller circle
The chord of contact from an external point \( (x_1, y_1) \) to the smaller circle \( C_1 \) is given by:
\[ x_1 x + y_1 y = a^2 \]
Substituting \( x_1 \) and \( y_1 \):
\[ (\sqrt{2}a \cos \theta)x + (\sqrt{2}a \sin \theta)y = a^2 \]
Dividing the entire equation by \( a \):
\[ \sqrt{2} \cos \theta \cdot x + \sqrt{2} \sin \theta \cdot y = a \]
This simplifies to:
\[ x \cos \theta + y \sin \theta = \frac{a}{\sqrt{2}} \]

### Step 4: Extend the chord of contact to meet the larger circle again
Let the chord of contact intersect the larger circle \( C_2 \) again at points \( A' \) and \( B' \). The equation of the chord of contact remains the same.

### Step 5: Find the tangents at points A' and B'
The tangents at points \( A' \) and \( B' \) from the point of intersection \( Q \) will also satisfy the chord of contact equation for the larger circle:
\[ x_1 x + y_1 y = 2a^2 \]
where \( (x_1, y_1) \) is the point of intersection of the tangents.

### Step 6: Set up the equations for locus
We have two equations:
1. From the smaller circle:
\[ x \cos \theta + y \sin \theta = \frac{a}{\sqrt{2}} \]
2. From the larger circle:
\[ x_1 x + y_1 y = 2a^2 \]

### Step 7: Compare coefficients to find the locus
Comparing the coefficients of \( x \) and \( y \) from both equations, we can derive relationships between \( x_1, y_1 \) and \( \theta \).

### Step 8: Use the identity \( \sin^2 \theta + \cos^2 \theta = 1 \)
From the relationships derived, we can express \( x_1 \) and \( y_1 \) in terms of \( \theta \):
\[ \cos \theta = \frac{x_1}{2\sqrt{2}a}, \quad \sin \theta = \frac{y_1}{2\sqrt{2}a} \]
Using the identity \( \sin^2 \theta + \cos^2 \theta = 1 \):
\[ \left(\frac{x_1}{2\sqrt{2}a}\right)^2 + \left(\frac{y_1}{2\sqrt{2}a}\right)^2 = 1 \]
This simplifies to:
\[ \frac{x_1^2 + y_1^2}{8a^2} = 1 \]
Thus, we have:
\[ x_1^2 + y_1^2 = 8a^2 \]

### Final Step: Write the locus equation
Replacing \( x_1 \) and \( y_1 \) with \( x \) and \( y \), we get the locus of the point of intersection of the tangents at points \( A' \) and \( B' \):
\[ x^2 + y^2 = 8a^2 \]

### Conclusion
The locus of the point of intersection of the tangents at points A' and B' is given by:
\[ \boxed{x^2 + y^2 = 8a^2} \]