Find the polar of the point `(5, - (1)/(2))` with respect to the circle ` 3 x^(2) + 3y ^(2) - 7 x + 8y - 9 = 0`

Answer

To find the polar of the point \((5, -\frac{1}{2})\) with respect to the circle given by the equation \(3x^2 + 3y^2 - 7x + 8y - 9 = 0\), 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:
\[ 3x^2 + 3y^2 - 7x + 8y - 9 = 0 \]
Dividing the entire equation by 3:
\[ x^2 + y^2 - \frac{7}{3}x + \frac{8}{3}y - 3 = 0 \]

### Step 2: Identify the Coefficients
From the standard form of a circle \(x^2 + y^2 + 2gx + 2fy + c = 0\), we can identify:
- \(g = -\frac{7}{6}\)
- \(f = \frac{4}{3}\)
- \(c = -3\)

### Step 3: Use the Polar Equation Formula
The polar of a point \((x_1, y_1)\) with respect to a circle can be found using the formula:
\[ xx_1 + yy_1 + g(x + x_1) + f(y + y_1) + c = 0 \]
Here, \((x_1, y_1) = (5, -\frac{1}{2})\).

### Step 4: Substitute Values into the Polar Equation
Substituting \(x_1 = 5\) and \(y_1 = -\frac{1}{2}\) into the polar equation:
\[ x(5) + y\left(-\frac{1}{2}\right) + \left(-\frac{7}{6}\right)(x + 5) + \left(\frac{4}{3}\right)\left(y - \frac{1}{2}\right) - 3 = 0 \]

### Step 5: Simplify the Equation
Now, we simplify the equation:
1. The first term: \(5x\)
2. The second term: \(-\frac{1}{2}y\)
3. The third term: \(-\frac{7}{6}x - \frac{35}{6}\)
4. The fourth term: \(\frac{4}{3}y - \frac{2}{3}\)

Combining these:
\[ 5x - \frac{7}{6}x + \frac{4}{3}y - \frac{1}{2}y - \frac{35}{6} - \frac{2}{3} - 3 = 0 \]

### Step 6: Combine Like Terms
Combining the \(x\) terms:
\[ \left(5 - \frac{7}{6}\right)x = \left(\frac{30}{6} - \frac{7}{6}\right)x = \frac{23}{6}x \]
Combining the \(y\) terms:
\[ \left(\frac{4}{3} - \frac{1}{2}\right)y = \left(\frac{8}{6} - \frac{3}{6}\right)y = \frac{5}{6}y \]
Combining the constant terms:
\[ -\frac{35}{6} - \frac{4}{6} - 3 = -\frac{35 + 4 + 18}{6} = -\frac{57}{6} \]

### Final Polar Equation
Putting it all together, we have:
\[ \frac{23}{6}x + \frac{5}{6}y - \frac{57}{6} = 0 \]
Multiplying through by 6 to eliminate the fractions:
\[ 23x + 5y - 57 = 0 \]

### Solution
Thus, the polar of the point \((5, -\frac{1}{2})\) with respect to the given circle is:
\[ 23x + 5y - 57 = 0 \]