The equation of the circle which passes through `(2a, 0)` and has the radical axis `2x - a =0` withthe circie `x^2+y^2-a^2=0` is

Options

A. `x^(2) + y^(2) - ax = 0`
B. `x^(2) + y^(2) + 2ax = 0`
C. `x^(2) + y^(2) -2ax =0`
D. `x^(2) + y^(2) + ax = 0`

Answer

To find the equation of the circle that passes through the point \((2a, 0)\) and has the radical axis \(2x - a = 0\) with respect to the circle defined by \(x^2 + y^2 - a^2 = 0\), we can follow these steps:

### Step 1: Understand the given information
We have:
- A point through which the circle passes: \((2a, 0)\)
- The radical axis: \(2x - a = 0\) (which simplifies to \(x = \frac{a}{2}\))
- The equation of the reference circle: \(x^2 + y^2 - a^2 = 0\)

### Step 2: Define the general equation of the circle
The general equation of a circle can be written as:
\[ S_1 = x^2 + y^2 + 2gx + 2fy + c = 0 \]
where \((g, f)\) are the coordinates of the center and \(c\) is a constant.

### Step 3: Set up the equation based on the radical axis
The radical axis condition states that:
\[ S_1 - S_2 = 0 \]
where \(S_2 = x^2 + y^2 - a^2 = 0\).

Thus, we can write:
\[ S_1 - (x^2 + y^2 - a^2) = 0 \]
This simplifies to:
\[ S_1 = x^2 + y^2 - a^2 \]
Substituting \(S_1\):
\[ x^2 + y^2 + 2gx + 2fy + c - (x^2 + y^2 - a^2) = 0 \]
This leads to:
\[ 2gx + 2fy + c + a^2 = 0 \]

### Step 4: Use the point \((2a, 0)\)
Since the circle passes through the point \((2a, 0)\), we substitute \(x = 2a\) and \(y = 0\) into the equation:
\[ 2g(2a) + 2f(0) + c + a^2 = 0 \]
This simplifies to:
\[ 4ag + c + a^2 = 0 \quad \text{(Equation 1)} \]

### Step 5: Analyze the radical axis
The radical axis \(2x - a = 0\) suggests that the center of the circle lies on this line. Therefore, we can express \(g\) in terms of \(f\):
\[ g = \frac{a}{4} \]

### Step 6: Substitute \(g\) into Equation 1
Substituting \(g = \frac{a}{4}\) into Equation 1:
\[ 4a\left(\frac{a}{4}\right) + c + a^2 = 0 \]
This simplifies to:
\[ a^2 + c + a^2 = 0 \implies 2a^2 + c = 0 \implies c = -2a^2 \]

### Step 7: Write the final equation of the circle
Substituting \(g\) and \(c\) back into the general equation of the circle:
\[ x^2 + y^2 + 2\left(\frac{a}{4}\right)x + 2fy - 2a^2 = 0 \]
This gives us:
\[ x^2 + y^2 + \frac{a}{2}x + 2fy - 2a^2 = 0 \]

### Step 8: Choose \(f\)
Since \(f\) is not determined by the previous equations, we can set \(f = 0\) for simplicity. Thus, the equation becomes:
\[ x^2 + y^2 + \frac{a}{2}x - 2a^2 = 0 \]

### Final Equation
The equation of the circle is:
\[ x^2 + y^2 - \frac{a}{2}x - 2a^2 = 0 \]