Find the locus of the point whose polars with respect to the circles `x^(2)+y^(2)-4x-4y-8=0` and `x^(2)+y^(2)-2x+6y-2=0` are mutually perpendicular.

Answer

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

### Step 1: Identify the circles and their equations
The equations of the circles are:
1. \( C_1: x^2 + y^2 - 4x - 4y - 8 = 0 \)
2. \( C_2: x^2 + y^2 - 2x + 6y - 2 = 0 \)

### Step 2: Rewrite the equations in standard form
We can rewrite the equations of the circles in standard form by completing the square.

For \( C_1 \):
\[ x^2 - 4x + y^2 - 4y = 8 \]
Completing the square:
\[ (x-2)^2 + (y-2)^2 = 12 \]
This represents a circle centered at \( (2, 2) \) with radius \( 2\sqrt{3} \).

For \( C_2 \):
\[ x^2 - 2x + y^2 + 6y = 2 \]
Completing the square:
\[ (x-1)^2 + (y+3)^2 = 12 \]
This represents a circle centered at \( (1, -3) \) with radius \( 2\sqrt{3} \).

### Step 3: Find the polar equations
Let \( P(x_1, y_1) \) be a point. The polar of point \( P \) with respect to circle \( C_1 \) is given by:
\[ xx_1 + yy_1 - 2(x + 2) - 2(y + 2) - 8 = 0 \]
Simplifying gives:
\[ xx_1 + yy_1 - 2x - 2y - 2x_1 - 2y_1 - 8 = 0 \]

The polar of point \( P \) with respect to circle \( C_2 \) is given by:
\[ xx_1 + yy_1 - (x + 1) - (y - 3) - 2 = 0 \]
Simplifying gives:
\[ xx_1 + yy_1 - x - y - x_1 + 3y_1 - 2 = 0 \]

### Step 4: Set up the perpendicular condition
The two polars are mutually perpendicular if:
\[ a_1 a_2 + b_1 b_2 = 0 \]
where \( a_1, b_1 \) are the coefficients of \( x \) and \( y \) in the first polar, and \( a_2, b_2 \) are the coefficients in the second polar.

From the polar equations, we can identify:
1. For the first polar: \( a_1 = x_1 - 2 \), \( b_1 = y_1 - 2 \)
2. For the second polar: \( a_2 = x_1 - 1 \), \( b_2 = y_1 + 3 \)

Setting up the condition:
\[ (x_1 - 2)(x_1 - 1) + (y_1 - 2)(y_1 + 3) = 0 \]

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

### Step 6: Replace \( x_1, y_1 \) with \( x, y \)
To find the locus, replace \( x_1 \) and \( y_1 \) with \( x \) and \( y \):
\[ x^2 + y^2 - 3x + y - 4 = 0 \]

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