Find the pole of the straight line
`2 x - y = 6` with respect to the circle ` 5 x^(2) + 5y ^(2) = 9`

Answer

To find the pole of the straight line \(2x - y = 6\) with respect to the circle \(5x^2 + 5y^2 = 9\), we will follow these steps:

### Step 1: Rewrite the Circle Equation
First, we rewrite the equation of the circle in standard form:
\[ 5x^2 + 5y^2 = 9 \implies x^2 + y^2 = \frac{9}{5} \]
This means the circle is centered at the origin with a radius of \(\sqrt{\frac{9}{5}}\).

### Step 2: Identify the Line Equation
The given line equation is:
\[ 2x - y - 6 = 0 \]
This can be rewritten as:
\[ y = 2x - 6 \]

### Step 3: Polar Equation of the Circle
For the circle \(x^2 + y^2 + 2gx + 2fy + c = 0\), we have:
- \(g = 0\)
- \(f = 0\)
- \(c = -\frac{9}{5}\)

The polar equation with respect to a point \((\alpha, \beta)\) is given by:
\[ x\alpha + y\beta + \frac{9}{5} = 0 \]

### Step 4: Set Up the Polar Equation
Since the line \(2x - y - 6 = 0\) is the polar of the point \((\alpha, \beta)\), we can equate the coefficients of the polar equation to that of the line:
\[ \frac{9}{5} = -6 \quad \text{(constant term)} \]
\[ \beta = -1 \quad \text{(coefficient of } y\text{)} \]
\[ 2 = \alpha \quad \text{(coefficient of } x\text{)} \]

### Step 5: Solve for \(\alpha\) and \(\beta\)
From the equations, we can solve for \(\alpha\) and \(\beta\):
1. From the coefficient of \(x\):
\[ 2 = 5\alpha \implies \alpha = \frac{2}{5} \]

2. From the coefficient of \(y\):
\[ -1 = 5\beta \implies \beta = -\frac{1}{5} \]

### Step 6: Conclusion
Thus, the pole of the line \(2x - y = 6\) with respect to the circle \(5x^2 + 5y^2 = 9\) is:
\[ \left(\frac{2}{5}, -\frac{1}{5}\right) \]