The normal at the point (3, 4) on a circle cuts the circle at the point (-1,-2). Then the equation of the circle is

Options

A. `x^(2) +y^(2)+2x -2y - 13 = 0`
B. `x^(2) +y^(2) -2x -2y -11 = 0`
C. `x^(2)+y^(2)-2x +2y +12 = 0`
D. `x^(2) +y^(2) - 2x - 2y+14 = 0`

Explanation

To find the equation of the circle given the points (3, 4) and (-1, -2), we can follow these steps:

### Step 1: Identify the endpoints of the diameter
The points (3, 4) and (-1, -2) are the endpoints of the diameter of the circle.

### Step 2: Find the center of the circle
The center of the circle can be found by calculating the midpoint of the diameter. The midpoint \(M\) of the points \((x_1, y_1)\) and \((x_2, y_2)\) is given by:
\[ M = \left( \frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2} \right) \]
Substituting the values:
\[ M = \left( \frac{3 + (-1)}{2}, \frac{4 + (-2)}{2} \right) = \left( \frac{2}{2}, \frac{2}{2} \right) = (1, 1) \]

### Step 3: Calculate the radius of the circle
The radius \(r\) of the circle is half the distance between the two endpoints. We can find the distance \(d\) between the points using the distance formula:
\[ d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \]
Substituting the values:
\[ d = \sqrt{((-1) - 3)^2 + ((-2) - 4)^2} = \sqrt{(-4)^2 + (-6)^2} = \sqrt{16 + 36} = \sqrt{52} = 2\sqrt{13} \]
Thus, the radius \(r\) is:
\[ r = \frac{d}{2} = \frac{2\sqrt{13}}{2} = \sqrt{13} \]

### Step 4: Write the equation of the circle
The standard form of the equation of a circle with center \((h, k)\) and radius \(r\) is:
\[ (x - h)^2 + (y - k)^2 = r^2 \]
Substituting \(h = 1\), \(k = 1\), and \(r^2 = 13\):
\[ (x - 1)^2 + (y - 1)^2 = 13 \]

### Step 5: Expand the equation
Expanding the equation:
\[ (x - 1)^2 + (y - 1)^2 = 13 \]
\[ x^2 - 2x + 1 + y^2 - 2y + 1 = 13 \]
\[ x^2 + y^2 - 2x - 2y + 2 = 13 \]
\[ x^2 + y^2 - 2x - 2y - 11 = 0 \]

### Final Equation
Thus, the equation of the circle is:
\[ x^2 + y^2 - 2x - 2y - 11 = 0 \]

### Conclusion
The correct option is option 2:
\[ x^2 + y^2 - 2x - 2y - 11 = 0 \]