The inverse of the point (1, 2) with respect to the circle `x^(2) + y^(2) - 4x - 6y + 9 = 0` is

Options

A. (0,0)
B. (1,0)
C. (0,1)
D. (1,1)

Answer

To find the inverse of the point (1, 2) with respect to the circle given by the equation \(x^2 + y^2 - 4x - 6y + 9 = 0\), 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:

\[ x^2 + y^2 - 4x - 6y + 9 = 0 \]

We can rearrange this to group the \(x\) and \(y\) terms:

\[ x^2 - 4x + y^2 - 6y + 9 = 0 \]

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

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

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

Substituting these back into the equation gives:

\[ (x - 2)^2 - 4 + (y - 3)^2 - 9 + 9 = 0 \]

This simplifies to:

\[ (x - 2)^2 + (y - 3)^2 - 4 = 0 \]

Thus, we can write:

\[ (x - 2)^2 + (y - 3)^2 = 4 \]

This shows that the circle has a center at \((2, 3)\) and a radius of \(2\).

### Step 3: Find the Inverse Point
To find the inverse of the point \((1, 2)\) with respect to the circle, we will use the formula for the inverse point \((x', y')\) which is given by:

\[ x' = \frac{r^2 (x - h)}{(x - h)^2 + (y - k)^2} + h \]
\[ y' = \frac{r^2 (y - k)}{(x - h)^2 + (y - k)^2} + k \]

Where \((h, k)\) is the center of the circle, and \(r\) is the radius.

Here, \(h = 2\), \(k = 3\), and \(r^2 = 4\).

Substituting the values for the point \((1, 2)\):

1. Calculate \((x - h)\) and \((y - k)\):
- \(x - h = 1 - 2 = -1\)
- \(y - k = 2 - 3 = -1\)

2. Calculate \((x - h)^2 + (y - k)^2\):
- \((-1)^2 + (-1)^2 = 1 + 1 = 2\)

3. Now substitute into the inverse formulas:
- \(x' = \frac{4 \cdot (-1)}{2} + 2 = \frac{-4}{2} + 2 = -2 + 2 = 0\)
- \(y' = \frac{4 \cdot (-1)}{2} + 3 = \frac{-4}{2} + 3 = -2 + 3 = 1\)

Thus, the inverse of the point \((1, 2)\) with respect to the circle is \((0, 1)\).

### Final Answer
The inverse of the point (1, 2) with respect to the circle is \((0, 1)\).