Find the equation of the normal to the
circle `x^(2) + y^(2)- 4x-6y+11=0` at `(3,2)`.
Also find the other point where the
normal meets the circle.

Answer

To find the equation of the normal to the circle given by the equation \(x^2 + y^2 - 4x - 6y + 11 = 0\) at the point \((3, 2)\), and also to find the other point where the normal meets the circle, we can follow these steps:

### Step 1: Rewrite the Equation of the Circle
First, let's rewrite the equation of the circle in a more standard form. We can complete the square for both \(x\) and \(y\).

The given equation is:
\[ x^2 - 4x + y^2 - 6y + 11 = 0 \]

Completing the square for \(x\):
\[ x^2 - 4x = (x - 2)^2 - 4 \]

Completing the square for \(y\):
\[ y^2 - 6y = (y - 3)^2 - 9 \]

Substituting these into the equation gives:
\[ (x - 2)^2 - 4 + (y - 3)^2 - 9 + 11 = 0 \]
\[ (x - 2)^2 + (y - 3)^2 - 2 = 0 \]
\[ (x - 2)^2 + (y - 3)^2 = 2 \]

This represents a circle centered at \((2, 3)\) with a radius of \(\sqrt{2}\).

### Step 2: Find the Slope of the Tangent at the Point (3, 2)
To find the slope of the normal, we first need to find the slope of the tangent line at the point \((3, 2)\). We can do this by differentiating the equation of the circle implicitly.

Differentiating the equation \(x^2 + y^2 - 4x - 6y + 11 = 0\):
\[ 2x + 2y \frac{dy}{dx} - 4 - 6 \frac{dy}{dx} = 0 \]
Rearranging gives:
\[ (2y - 6) \frac{dy}{dx} = 4 - 2x \]
\[ \frac{dy}{dx} = \frac{4 - 2x}{2y - 6} \]

Now substituting \(x = 3\) and \(y = 2\):
\[ \frac{dy}{dx} = \frac{4 - 2(3)}{2(2) - 6} = \frac{4 - 6}{4 - 6} = \frac{-2}{-2} = 1 \]

Thus, the slope of the tangent at \((3, 2)\) is \(1\).

### Step 3: Find the Slope of the Normal
The slope of the normal is the negative reciprocal of the slope of the tangent:
\[ \text{slope of normal} = -\frac{1}{1} = -1 \]

### Step 4: Write the Equation of the Normal
Using the point-slope form of the equation of a line:
\[ y - y_0 = m(x - x_0) \]
where \((x_0, y_0) = (3, 2)\) and \(m = -1\):
\[ y - 2 = -1(x - 3) \]
Simplifying this gives:
\[ y - 2 = -x + 3 \]
\[ x + y = 5 \]

### Step 5: Find the Other Point Where the Normal Meets the Circle
We need to find the intersection of the line \(x + y = 5\) with the circle \((x - 2)^2 + (y - 3)^2 = 2\).

Substituting \(y = 5 - x\) into the circle's equation:
\[ (x - 2)^2 + ((5 - x) - 3)^2 = 2 \]
\[ (x - 2)^2 + (2 - x)^2 = 2 \]
Expanding both squares:
\[ (x - 2)^2 = x^2 - 4x + 4 \]
\[ (2 - x)^2 = x^2 - 4x + 4 \]
Combining gives:
\[ x^2 - 4x + 4 + x^2 - 4x + 4 = 2 \]
\[ 2x^2 - 8x + 8 = 2 \]
\[ 2x^2 - 8x + 6 = 0 \]
Dividing by 2:
\[ x^2 - 4x + 3 = 0 \]
Factoring:
\[ (x - 3)(x - 1) = 0 \]
Thus, \(x = 3\) or \(x = 1\).

Since we already have the point \((3, 2)\), we take \(x = 1\):
Substituting \(x = 1\) back into \(y = 5 - x\):
\[ y = 5 - 1 = 4 \]
Thus, the other point where the normal meets the circle is \((1, 4)\).

### Final Answers
- The equation of the normal is \(x + y = 5\).
- The other point where the normal meets the circle is \((1, 4)\).