Inverse of (0,0) w.r.t to circle `x^(2)+y^(2)-4x-6y+3=0` is

Options

A. `(6/13,9/13)`
B. `((-5)/13,9/13)`
C. `(6/13,(-9)/13)`
D. `((-6)/13,(-9)/13)`

Answer

To find the inverse of the point (0, 0) with respect to the circle given by the equation \( x^2 + y^2 - 4x - 6y + 3 = 0 \), 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:
\[ x^2 + y^2 - 4x - 6y + 3 = 0 \]
We can rearrange it to:
\[ x^2 - 4x + y^2 - 6y + 3 = 0 \]

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

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

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

Substituting these back into the equation gives:
\[ (x - 2)^2 - 4 + (y - 3)^2 - 9 + 3 = 0 \]
Simplifying this:
\[ (x - 2)^2 + (y - 3)^2 - 10 = 0 \]
Thus, we have:
\[ (x - 2)^2 + (y - 3)^2 = 10 \]
This shows that the center of the circle is at \( (2, 3) \) and the radius \( r \) is \( \sqrt{10} \).

### Step 3: Find the Value of \( \alpha \)
The formula for \( \alpha \) is given by:
\[ \alpha = \frac{r^2}{(x - h)^2 + (y - k)^2} \]
where \( (h, k) \) is the center of the circle. Here, \( r^2 = 10 \), \( h = 2 \), and \( k = 3 \).

We need to calculate \( (0 - 2)^2 + (0 - 3)^2 \):
\[ (0 - 2)^2 + (0 - 3)^2 = 4 + 9 = 13 \]

Now substituting into the formula for \( \alpha \):
\[ \alpha = \frac{10}{13} \]

### Step 4: Calculate the Inverse Point
The inverse point \( (x', y') \) can be calculated using the formulas:
\[ x' = \alpha (x - h) + h \]
\[ y' = \alpha (y - k) + k \]

Substituting \( x = 0 \), \( y = 0 \), \( h = 2 \), and \( k = 3 \):
\[ x' = \frac{10}{13} (0 - 2) + 2 = \frac{10}{13} (-2) + 2 = -\frac{20}{13} + \frac{26}{13} = \frac{6}{13} \]
\[ y' = \frac{10}{13} (0 - 3) + 3 = \frac{10}{13} (-3) + 3 = -\frac{30}{13} + \frac{39}{13} = \frac{9}{13} \]

Thus, the inverse point is:
\[ \left( \frac{6}{13}, \frac{9}{13} \right) \]

### Final Answer
The inverse of the point \( (0, 0) \) with respect to the circle is:
\[ \left( \frac{6}{13}, \frac{9}{13} \right) \]