Find the polar of the point (-2,3) with respect to the circle `x^2+y^2−4x−6y+5=0`

Answer

To find the polar of the point (-2, 3) with respect to the circle given by the equation \(x^2 + y^2 - 4x - 6y + 5 = 0\), we will follow these steps:

### Step 1: Rewrite the Circle Equation
First, we need to rewrite the equation of the circle in standard form. The given equation is:
\[ x^2 + y^2 - 4x - 6y + 5 = 0 \]
We can rearrange it as:
\[ x^2 - 4x + y^2 - 6y + 5 = 0 \]

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

For \(x^2 - 4x\):
\[ x^2 - 4x = (x - 2)^2 - 4 \]

For \(y^2 - 6y\):
\[ y^2 - 6y = (y - 3)^2 - 9 \]

Now substituting back into the equation:
\[ (x - 2)^2 - 4 + (y - 3)^2 - 9 + 5 = 0 \]
This simplifies to:
\[ (x - 2)^2 + (y - 3)^2 - 8 = 0 \]
or
\[ (x - 2)^2 + (y - 3)^2 = 8 \]
This represents a circle with center (2, 3) and radius \( \sqrt{8} \).

### Step 3: Identify the Parameters
From the general equation of the circle, we have:
- \( g = -4 \)
- \( f = -6 \)
- \( c = 5 \)

### Step 4: Use the Polar Equation Formula
The polar of a point \((x_1, y_1)\) with respect to a circle defined by \(x^2 + y^2 + gx + fy + c = 0\) is given by:
\[ xx_1 + yy_1 + \frac{g}{2}(x + x_1) + \frac{f}{2}(y + y_1) + c = 0 \]
Substituting \((x_1, y_1) = (-2, 3)\):
\[ x(-2) + y(3) + \frac{-4}{2}(x - 2) + \frac{-6}{2}(y + 3) + 5 = 0 \]

### Step 5: Simplify the Equation
Now, substituting the values:
\[ -2x + 3y - 2(x - 2) - 3(y + 3) + 5 = 0 \]
This expands to:
\[ -2x + 3y - 2x + 4 - 3y - 9 + 5 = 0 \]
Combining like terms:
\[ -4x + 0y + 0 = 0 \]
This simplifies to:
\[ -4x = 0 \]
or
\[ x = 0 \]

### Conclusion
Thus, the polar of the point (-2, 3) with respect to the given circle is:
\[ \boxed{x = 0} \]