Find the pole of the straight line
`48 x - 5y + 53 = 0` with respect to the circle ` 3x ^(2) + 3y ^(2) + 5x - 7y + 2 = 0`

Answer

To find the pole of the straight line \( 48x - 5y + 53 = 0 \) with respect to the circle \( 3x^2 + 3y^2 + 5x - 7y + 2 = 0 \), we will follow these steps:

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

### Step 2: Complete the Square
Next, we complete the square for both \(x\) and \(y\).

For \(x\):
\[ x^2 + \frac{5}{3}x = \left(x + \frac{5}{6}\right)^2 - \frac{25}{36} \]

For \(y\):
\[ y^2 - \frac{7}{3}y = \left(y - \frac{7}{6}\right)^2 - \frac{49}{36} \]

Substituting these back into the circle equation gives:
\[ \left(x + \frac{5}{6}\right)^2 - \frac{25}{36} + \left(y - \frac{7}{6}\right)^2 - \frac{49}{36} + \frac{2}{3} = 0 \]

### Step 3: Simplify the Circle Equation
Now, we simplify:
\[ \left(x + \frac{5}{6}\right)^2 + \left(y - \frac{7}{6}\right)^2 - \frac{25 + 49 - 24}{36} = 0 \]
This simplifies to:
\[ \left(x + \frac{5}{6}\right)^2 + \left(y - \frac{7}{6}\right)^2 = \frac{2}{36} = \frac{1}{18} \]
This shows that the center of the circle is at \( \left(-\frac{5}{6}, \frac{7}{6}\right) \) and the radius is \( \frac{1}{\sqrt{18}} \).

### Step 4: Find the Pole of the Line
The pole of a line \(Ax + By + C = 0\) with respect to a circle \(x^2 + y^2 + Dx + Ey + F = 0\) is given by the formula:
\[ \left(-\frac{A}{D}, -\frac{B}{E}\right) \]
For our line \(48x - 5y + 53 = 0\), we have \(A = 48\), \(B = -5\), and \(C = 53\). For the circle, we can rewrite it as:
\[ x^2 + y^2 + \frac{5}{3}x - \frac{7}{3}y + \frac{2}{3} = 0 \]
So \(D = \frac{5}{3}\) and \(E = -\frac{7}{3}\).

Now we can find the coordinates of the pole:
\[ \text{Pole} = \left(-\frac{48}{\frac{5}{3}}, -\frac{-5}{-\frac{7}{3}}\right) \]
Calculating these gives:
\[ \text{Pole} = \left(-\frac{48 \cdot 3}{5}, -\frac{5 \cdot 3}{-7}\right) = \left(-\frac{144}{5}, \frac{15}{7}\right) \]

### Final Answer
Thus, the pole of the line \(48x - 5y + 53 = 0\) with respect to the circle \(3x^2 + 3y^2 + 5x - 7y + 2 = 0\) is:
\[ \left(-\frac{144}{5}, \frac{15}{7}\right) \]