Find the pole of the straight line ` x + 2 y = 1 ` with respect to the circle ` x^(2) + y^(2) = 5`

Answer

To find the pole of the straight line \( x + 2y = 1 \) with respect to the circle \( x^2 + y^2 = 5 \), we can follow these steps:

### Step 1: Write the equations in standard form
The equation of the circle can be rewritten as:
\[ x^2 + y^2 - 5 = 0 \]
This is in the form \( S = 0 \).

The equation of the line can be rewritten as:
\[ x + 2y - 1 = 0 \]
This is in the form \( L = 0 \).

### Step 2: Identify the coefficients
From the circle equation \( S = x^2 + y^2 - 5 \), we can identify:
- \( g = 0 \)
- \( f = 0 \)
- \( c = -5 \)

From the line equation \( L = x + 2y - 1 \), we can identify:
- \( A = 1 \)
- \( B = 2 \)
- \( C = -1 \)

### Step 3: Use the formula for the pole
The pole \( (x_1, y_1) \) of the line with respect to the circle can be found using the formulas:
\[ \alpha = -\frac{g}{A} \quad \text{and} \quad \beta = -\frac{f}{B} \]

### Step 4: Substitute the values
Substituting the values we found:
- For \( \alpha \):
\[ \alpha = -\frac{0}{1} = 0 \]
- For \( \beta \):
\[ \beta = -\frac{0}{2} = 0 \]

### Step 5: Calculate the pole
Now we need to find the constant term:
\[ C = -1 \quad \text{and} \quad c = -5 \]
Using the relation:
\[ \frac{C}{c} = \frac{-1}{-5} = \frac{1}{5} \]
This gives us the coordinates of the pole:
\[ \alpha = 5 \quad \text{and} \quad \beta = 10 \]

### Final Result
Thus, the pole of the line \( x + 2y = 1 \) with respect to the circle \( x^2 + y^2 = 5 \) is:
\[ (5, 10) \]

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