Find the pole of the straight line
` 9 x + y - 28 = 0 …………(1)`
with respect to the circle
` 2 x^(2) + 2y^(2) - 3x + 5y - 7 = 0 ………….(2)`

Answer

To find the pole of the straight line \( 9x + y - 28 = 0 \) with respect to the circle \( 2x^2 + 2y^2 - 3x + 5y - 7 = 0 \), we follow these steps:

### Step 1: Rewrite the circle equation in standard form
The given circle equation is:
\[ 2x^2 + 2y^2 - 3x + 5y - 7 = 0 \]
We can divide the entire equation by 2 to simplify it:
\[ x^2 + y^2 - \frac{3}{2}x + \frac{5}{2}y - \frac{7}{2} = 0 \]

### Step 2: Identify the coefficients
The standard form of a circle is:
\[ x^2 + y^2 + 2gx + 2fy + c = 0 \]
From our equation, we can identify:
- \( 2g = -\frac{3}{2} \) → \( g = -\frac{3}{4} \)
- \( 2f = \frac{5}{2} \) → \( f = \frac{5}{4} \)
- \( c = -\frac{7}{2} \)

### Step 3: Set up the polar equation
The polar equation of a point \( (h, k) \) with respect to the circle is given by:
\[ xh + yk + g(x^2 + y^2) + f(x + y) + c = 0 \]
Substituting \( g \), \( f \), and \( c \) into the polar equation, we have:
\[ xh + yk - \frac{3}{4}(x^2 + y^2) + \frac{5}{4}(x + y) - \frac{7}{2} = 0 \]

### Step 4: Substitute the line equation
We need to express the line \( 9x + y - 28 = 0 \) in terms of \( h \) and \( k \):
\[ 9x + y = 28 \]
This means \( y = 28 - 9x \).

### Step 5: Equate coefficients
We can equate the coefficients of \( x \) and \( y \) from the polar equation and the line equation. The polar equation can be simplified to:
\[ xh + yk - \frac{3}{4}x^2 - \frac{3}{4}y^2 + \frac{5}{4}x + \frac{5}{4}y - \frac{7}{2} = 0 \]
Comparing coefficients with \( 9x + y - 28 = 0 \):
1. Coefficient of \( x \): \( h - \frac{3}{4} = 9 \)
2. Coefficient of \( y \): \( k + \frac{5}{4} = 1 \)
3. Constant term: \( -\frac{7}{2} = -28 \)

### Step 6: Solve for \( h \) and \( k \)
From the first equation:
\[ h - \frac{3}{4} = 9 \implies h = 9 + \frac{3}{4} = \frac{36}{4} + \frac{3}{4} = \frac{39}{4} \]

From the second equation:
\[ k + \frac{5}{4} = 1 \implies k = 1 - \frac{5}{4} = \frac{4}{4} - \frac{5}{4} = -\frac{1}{4} \]

### Step 7: Final result
Thus, the coordinates of the pole \( (h, k) \) are:
\[ \left( \frac{39}{4}, -\frac{1}{4} \right) \]