The locus of poles of tangents to the circle `x^(2)+y^(2)=a^(2)` w.r.t the circle `x^(2)+y^(2)+2ax-a^(2)=0` is

Options

A. `y^(2)=4ax`
B. `y^(2)=2ax`
C. `y^(2)+2ax=0`
D. `y^(2)+4ax=0`

Answer

To find the locus of the poles of tangents to the circle \( x^2 + y^2 = a^2 \) with respect to the circle \( x^2 + y^2 + 2ax - a^2 = 0 \), we can follow these steps:

### Step 1: Identify the equations of the circles
The first circle is given by:
\[ x^2 + y^2 = a^2 \]
The second circle can be rewritten as:
\[ x^2 + y^2 + 2ax - a^2 = 0 \implies x^2 + y^2 = a^2 - 2ax \]

### Step 2: Write the equation of the tangent to the first circle
The equation of the tangent to the circle \( x^2 + y^2 = a^2 \) at the point \( (x_1, y_1) \) is given by:
\[ xx_1 + yy_1 = a^2 \]

### Step 3: Determine the pole of the tangent with respect to the second circle
Let the coordinates of the pole be \( (h, k) \). The equation of the pole with respect to the second circle is:
\[ xh + yk + ax + ay - a^2 = 0 \]
This can be rearranged to:
\[ (h + 2a)x + ky - a^2 = 0 \]

### Step 4: Compare coefficients
From the tangent equation \( xx_1 + yy_1 = a^2 \) and the pole equation \( (h + 2a)x + ky - a^2 = 0 \), we can compare coefficients:
1. Coefficient of \( x \): \( h + 2a = x_1 \)
2. Coefficient of \( y \): \( k = y_1 \)

### Step 5: Express \( x_1 \) and \( y_1 \) in terms of \( h \) and \( k \)
From the above equations, we have:
\[ x_1 = h + 2a \quad \text{and} \quad y_1 = k \]

### Step 6: Substitute into the circle equation
Since \( (x_1, y_1) \) lies on the circle \( x^2 + y^2 = a^2 \), we substitute:
\[ (h + 2a)^2 + k^2 = a^2 \]

### Step 7: Expand and simplify
Expanding the left side gives:
\[ (h^2 + 4ah + 4a^2 + k^2) = a^2 \]
Rearranging gives:
\[ h^2 + k^2 + 4ah + 4a^2 - a^2 = 0 \implies h^2 + k^2 + 4ah + 3a^2 = 0 \]

### Step 8: Replace \( h \) and \( k \) with \( x \) and \( y \)
Let \( h = x \) and \( k = y \):
\[ x^2 + y^2 + 4ax + 3a^2 = 0 \]

### Step 9: Rearranging to find the locus
Rearranging gives:
\[ x^2 + y^2 = -4ax - 3a^2 \]

### Step 10: Identify the locus
This equation represents a parabola. Thus, the locus of the poles of the tangents to the circle \( x^2 + y^2 = a^2 \) with respect to the circle \( x^2 + y^2 + 2ax - a^2 = 0 \) is:
\[ y^2 = -4ax - 3a^2 \]

### Final Answer
The locus of poles of tangents to the circle \( x^2 + y^2 = a^2 \) with respect to the circle \( x^2 + y^2 + 2ax - a^2 = 0 \) is a parabola given by:
\[ y^2 = -4ax - 3a^2 \]