The locus of the point whose polars w.r.t. the circle `x^(2)+y^(2)-4x-4y-8=0 and x^(2)+y^(2)-2x+6y-2=0` are mutually perpendicular is

Options

A. `x^(2)+y^(2)-3x+y=4`
B. `x^(2)+y^(2)+3x-y=4`
C. `x^(2)+y^(2)+3x+y=4`
D. `x^(2)+y^(2)-3x-y=4`.

Answer

To find the locus of the point whose polars with respect to the given circles are mutually perpendicular, we can follow these steps:

### Step 1: Write the equations of the circles
The given equations of the circles are:
1. \( x^2 + y^2 - 4x - 4y - 8 = 0 \)
2. \( x^2 + y^2 - 2x + 6y - 2 = 0 \)

### Step 2: Identify the polar equations
Let the point \( P \) have coordinates \( (x_1, y_1) \). The polar of point \( P \) with respect to the first circle can be derived from the general form of the polar equation.

For the first circle, the polar equation is:
\[ x_1 x + y_1 y - 2(x + 2) - 2(y - 2) - 8 = 0 \]
This simplifies to:
\[ x_1 x + y_1 y - 2x - 4 - 2y + 4 - 8 = 0 \]
\[ x_1 x + y_1 y - 2x - 2y - 8 = 0 \]

For the second circle, the polar equation is:
\[ x_1 x + y_1 y - (x + 1) + 3(y - 2) - 2 = 0 \]
This simplifies to:
\[ x_1 x + y_1 y - x - 1 + 3y - 6 - 2 = 0 \]
\[ x_1 x + y_1 y - x + 3y - 9 = 0 \]

### Step 3: Set up the condition for perpendicularity
The polars are mutually perpendicular if the following condition holds:
\[ a_1 a_2 + b_1 b_2 = 0 \]
where \( a_1, b_1 \) are coefficients of \( x \) and \( y \) from the first polar, and \( a_2, b_2 \) are coefficients from the second polar.

From the first polar:
- \( a_1 = x_1 - 2 \)
- \( b_1 = y_1 - 2 \)

From the second polar:
- \( a_2 = x_1 - 1 \)
- \( b_2 = y_1 + 3 \)

### Step 4: Substitute into the perpendicularity condition
Substituting into the condition gives:
\[ (x_1 - 2)(x_1 - 1) + (y_1 - 2)(y_1 + 3) = 0 \]

### Step 5: Expand and simplify
Expanding this:
\[ x_1^2 - 3x_1 + 2 + y_1^2 + y_1 - 6 = 0 \]
Combining like terms:
\[ x_1^2 + y_1^2 - 3x_1 + y_1 - 4 = 0 \]

### Step 6: Replace \( x_1 \) and \( y_1 \) with \( x \) and \( y \)
Since \( x_1 \) and \( y_1 \) represent the coordinates of the locus, we can replace them with \( x \) and \( y \):
\[ x^2 + y^2 - 3x + y - 4 = 0 \]

### Final Result
The locus of the point whose polars with respect to the given circles are mutually perpendicular is:
\[ x^2 + y^2 - 3x + y - 4 = 0 \]