The inverse point of (2,-3) w.r.t to circle `x^(2)+y^(2)+6x-4y-12=0` is

Options

A. `(1/2,1/2)`
B. `(-1/2,1/2)`
C. `(1/2,-1/2)`
D. `(-1/2,-1/2)`

Answer

To find the inverse point of the point \( P(2, -3) \) with respect to the circle given by the equation \( x^2 + y^2 + 6x - 4y - 12 = 0 \), we will follow these steps:

### Step 1: Rewrite the equation of the circle in standard form
The given equation of the circle is:
\[ x^2 + y^2 + 6x - 4y - 12 = 0 \]
We will complete the square for both \( x \) and \( y \).

1. For \( x \):
\[ x^2 + 6x \quad \text{can be rewritten as} \quad (x + 3)^2 - 9 \]

2. For \( y \):
\[ y^2 - 4y \quad \text{can be rewritten as} \quad (y - 2)^2 - 4 \]

Now substituting back into the equation:
\[ (x + 3)^2 - 9 + (y - 2)^2 - 4 - 12 = 0 \]
Simplifying gives:
\[ (x + 3)^2 + (y - 2)^2 - 25 = 0 \]
Thus, we can express it as:
\[ (x + 3)^2 + (y - 2)^2 = 25 \]
This shows that the center of the circle is \( (-3, 2) \) and the radius \( r = 5 \).

### Step 2: Calculate the distance \( CP \) from the center to point \( P \)
The center \( C \) of the circle is \( (-3, 2) \) and the point \( P \) is \( (2, -3) \).

Using the distance formula:
\[ CP = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \]
Substituting the coordinates:
\[ CP = \sqrt{(2 - (-3))^2 + (-3 - 2)^2} = \sqrt{(2 + 3)^2 + (-5)^2} = \sqrt{5^2 + (-5)^2} = \sqrt{25 + 25} = \sqrt{50} = 5\sqrt{2} \]

### Step 3: Use the inverse point formula
The relationship for the inverse point \( P' \) with respect to the circle is given by:
\[ CP \cdot CP' = r^2 \]
Where \( r^2 = 25 \). Therefore:
\[ CP' = \frac{r^2}{CP} = \frac{25}{5\sqrt{2}} = \frac{5}{\sqrt{2}} = \frac{5\sqrt{2}}{2} \]

### Step 4: Find the coordinates of the inverse point \( P' \)
Let the coordinates of the inverse point \( P' \) be \( (x_1, y_1) \). The formulas for the coordinates of the inverse point are:
\[ x_1 = \frac{r^2 (x + 3)}{CP^2} - 3 \]
\[ y_1 = \frac{r^2 (y - 2)}{CP^2} + 2 \]
Substituting \( P(2, -3) \):
1. For \( x_1 \):
\[ x_1 = \frac{25(2 + 3)}{50} - 3 = \frac{25 \cdot 5}{50} - 3 = \frac{125}{50} - 3 = \frac{125}{50} - \frac{150}{50} = -\frac{25}{50} = -\frac{1}{2} \]

2. For \( y_1 \):
\[ y_1 = \frac{25(-3 - 2)}{50} + 2 = \frac{25 \cdot (-5)}{50} + 2 = -\frac{125}{50} + 2 = -\frac{125}{50} + \frac{100}{50} = -\frac{25}{50} = -\frac{1}{2} \]

### Final Result
Thus, the coordinates of the inverse point \( P' \) are:
\[ P' \left(-\frac{1}{2}, -\frac{1}{2}\right) \]