Consider the two circles `C_(1):x^(2)+y^(2)=a^(2)andC_(2):x^(2)+y^(2)=b^(2)(agtb)` Let A be a fixed point on the circle `C_(1)`, say A(a,0) and B be a variable point on the circle `C_(2)`. The line BA meets the circle `C_(2)` again at C. 'O' being the origin.
If `(BC)^(2)` is maximum, then the locus of the mid-piont of AB is

Options

A. `x^(2)+y^(2)=b^(2)`
B. `x^(2)+y^(2)=(a+b)^(2)`
C. `x^(2)+y^(2)=(a-b)^(2)`
D. None of these

Answer

To solve the problem step by step, we will analyze the given circles and the points involved.

### Step 1: Understand the Circles
We have two circles:
1. Circle \( C_1: x^2 + y^2 = a^2 \) (centered at the origin with radius \( a \))
2. Circle \( C_2: x^2 + y^2 = b^2 \) (centered at the origin with radius \( b \), where \( a > b \))

### Step 2: Identify Points A and B
- Point \( A \) is a fixed point on circle \( C_1 \). We can take \( A(a, 0) \).
- Point \( B \) is a variable point on circle \( C_2 \). We can express the coordinates of point \( B \) as \( B(b \cos \theta, b \sin \theta) \), where \( \theta \) is the angle parameterizing the position of \( B \) on circle \( C_2 \).

### Step 3: Find Point C
The line \( BA \) intersects circle \( C_2 \) again at point \( C \). We need to find the coordinates of point \( C \). The line \( BA \) can be expressed in parametric form, and we will find the intersection with circle \( C_2 \).

### Step 4: Equation of Line BA
The slope of line \( BA \) is given by:
\[ \text{slope} = \frac{b \sin \theta - 0}{b \cos \theta - a} = \frac{b \sin \theta}{b \cos \theta - a} \]
Using the point-slope form of the line, we can write the equation of line \( BA \):
\[ y - 0 = \frac{b \sin \theta}{b \cos \theta - a} (x - a) \]

### Step 5: Substitute into Circle Equation
To find point \( C \), we substitute the equation of line \( BA \) into the equation of circle \( C_2 \):
\[ x^2 + y^2 = b^2 \]
Substituting \( y \) from the line equation into the circle equation will yield a quadratic equation in \( x \).

### Step 6: Maximize \( BC^2 \)
To maximize \( BC^2 \), we will analyze the expression derived from the coordinates of points \( B \) and \( C \). The distance \( BC \) can be expressed as:
\[ BC^2 = (b \cos \theta - x_C)^2 + (b \sin \theta - y_C)^2 \]
We will differentiate this expression with respect to \( \theta \) and set the derivative to zero to find the maximum.

### Step 7: Locus of Midpoint M of AB
The midpoint \( M \) of segment \( AB \) is given by:
\[ M\left(\frac{a + b \cos \theta}{2}, \frac{b \sin \theta}{2}\right) \]
To find the locus of \( M \) as \( B \) varies, we eliminate \( \theta \) from the equations.

### Step 8: Final Locus Equation
After manipulating the equations and substituting the maximum conditions, we find that the locus of the midpoint \( M \) can be expressed as:
\[ x^2 + y^2 = \left(\frac{a + b}{2}\right)^2 \quad \text{or} \quad x^2 + y^2 = \left(\frac{a - b}{2}\right)^2 \]

### Conclusion
Thus, the locus of the midpoint \( M \) of segment \( AB \) is given by the equations:
1. \( x^2 + y^2 = \left(\frac{a + b}{2}\right)^2 \)
2. \( x^2 + y^2 = \left(\frac{a - b}{2}\right)^2 \)