Two variable chords AB and BC of a circle `x^(2)+y^(2)=a^(2)` are such that `AB=BC=a`. M and N are the midpoints of AB and BC, respectively, such that the line joining MN intersects the circles at P and Q, where P is closer to AB and O is the center of the circle.
The locus of the points of intersection of tangents at A and C is

Options

A. `x^(2)+y^(2)=a^(2)`
B. `x^(2)+y^(2)=2a^(2)`
C. `x^(2)+y^(2)=4a^(2)`
D. `x^(2)+y^(2)=8a^(2)`

Explanation

To find the locus of the points of intersection of the tangents at points A and C on the circle defined by the equation \( x^2 + y^2 = a^2 \), we can follow these steps:

### Step 1: Understand the Circle and Chords
The circle is defined by the equation \( x^2 + y^2 = a^2 \). The chords AB and BC are of equal length \( a \).

### Step 2: Identify Midpoints
Let M and N be the midpoints of chords AB and BC, respectively. Since AB = BC = a, we can express the coordinates of points A, B, and C in terms of the angle subtended at the center.

### Step 3: Parametrize Points A and C
Assume the coordinates of point A are \( (a \cos \theta_1, a \sin \theta_1) \) and point C are \( (a \cos \theta_2, a \sin \theta_2) \). The angle between the radius to A and C can be derived from the properties of the circle.

### Step 4: Find the Midpoints M and N
The midpoints M and N can be calculated as:
- \( M = \left( \frac{a \cos \theta_1 + a \cos \theta_2}{2}, \frac{a \sin \theta_1 + a \sin \theta_2}{2} \right) \)
- \( N = \left( \frac{a \cos \theta_2 + a \cos \theta_3}{2}, \frac{a \sin \theta_2 + a \sin \theta_3}{2} \right) \)

### Step 5: Equation of the Line MN
The line joining M and N can be expressed in a parametric form or slope-intercept form depending on the coordinates of M and N.

### Step 6: Find Intersection Points P and Q
The line MN intersects the circle at points P and Q. We can substitute the equation of the line into the circle's equation to find these intersection points.

### Step 7: Tangent at Points A and C
The equations of the tangents at points A and C can be derived using the point-slope form of the tangent to a circle. The tangent at point A is given by:
\[ x \cos \theta_1 + y \sin \theta_1 = a \]
And similarly for point C:
\[ x \cos \theta_2 + y \sin \theta_2 = a \]

### Step 8: Intersection of Tangents
To find the locus of the intersection of the tangents at points A and C, we can solve the two equations simultaneously.

### Step 9: Derive the Locus Equation
After solving the equations, we can derive the locus equation. The result will lead to the equation of a circle.

### Final Step: Identify the Correct Option
From the derived locus equation, we can compare it with the given options:
1. \( x^2 + y^2 = a^2 \)
2. \( x^2 + y^2 = 2a^2 \)
3. \( x^2 + y^2 = 4a^2 \)
4. \( x^2 + y^2 = 8a^2 \)

The correct answer is \( x^2 + y^2 = 4a^2 \).