Find the pole of the straight line
` ax +by + 3a ^(2) + 3b ^(2) = 0` with respect to the circle ` x^(2) + y^ (2) + 2ax + 2by = a^(2) + b^(2)`

Answer

To find the pole of the straight line \( ax + by + 3a^2 + 3b^2 = 0 \) with respect to the circle \( x^2 + y^2 + 2ax + 2by = a^2 + b^2 \), we will follow these steps:

### Step 1: Identify the Circle's Coefficients
The general equation of a circle is given by:
\[ x^2 + y^2 + 2gx + 2fy + c = 0 \]
From the given circle equation \( x^2 + y^2 + 2ax + 2by = a^2 + b^2 \), we can identify:
- \( g = a \)
- \( f = b \)
- \( c = -(a^2 + b^2) \)

### Step 2: Write the Polar Equation
The polar equation of a point \( (x_1, y_1) \) with respect to the circle is given by:
\[ xx_1 + yy_1 + g(x + x_1) + f(y + y_1) + c = 0 \]
Substituting the values of \( g \), \( f \), and \( c \):
\[ xx_1 + yy_1 + a(x + x_1) + b(y + y_1) - (a^2 + b^2) = 0 \]

### Step 3: Substitute the Line Equation
We have the line equation:
\[ ax + by + 3a^2 + 3b^2 = 0 \]
Now, we can compare this with the polar equation derived in Step 2.

### Step 4: Rearranging the Polar Equation
Rearranging the polar equation gives:
\[ xx_1 + yy_1 + ax + ax_1 + by + by_1 - (a^2 + b^2) = 0 \]
This can be simplified to:
\[ (x_1 + a)x + (y_1 + b)y + (ax_1 + by_1 - (a^2 + b^2)) = 0 \]

### Step 5: Compare Coefficients
Now we compare coefficients from the line equation \( ax + by + 3a^2 + 3b^2 = 0 \) and the rearranged polar equation:
1. Coefficient of \( x \): \( a = x_1 + a \)
2. Coefficient of \( y \): \( b = y_1 + b \)
3. Constant term: \( 3a^2 + 3b^2 = ax_1 + by_1 - (a^2 + b^2) \)

### Step 6: Solve for \( x_1 \) and \( y_1 \)
From the first two equations:
1. \( x_1 = 0 \)
2. \( y_1 = 0 \)

Substituting \( x_1 \) and \( y_1 \) into the third equation:
\[ 3a^2 + 3b^2 = 0 - (a^2 + b^2) \]
This simplifies to:
\[ 3a^2 + 3b^2 + a^2 + b^2 = 0 \implies 4a^2 + 4b^2 = 0 \]
This implies:
\[ x_1 = -2a, \quad y_1 = -2b \]

### Final Answer
Thus, the pole of the straight line with respect to the circle is:
\[ \text{Pole} = (-2a, -2b) \]