A rectangle ABCD is inscribed in the circle `x^2+y^2+3x+12y+2=0` . If the co-ordinates of A and B are (3,-2) and (-2,0) then the other two vertices of the rectangle are

Options

A. (-6,-10)
B. (-1,-12)
C. (1,12)
D. (6,10)

Answer

To find the coordinates of the other two vertices of rectangle ABCD inscribed in the given circle, we will follow these steps:

### Step 1: Write the circle equation in standard form
The given equation of the circle is:
\[ x^2 + y^2 + 3x + 12y + 2 = 0 \]

To convert it into standard form, we will complete the square for both \(x\) and \(y\).

1. Rearranging the equation:
\[ x^2 + 3x + y^2 + 12y = -2 \]

2. Completing the square for \(x\):
\[ x^2 + 3x = (x + \frac{3}{2})^2 - \frac{9}{4} \]

3. Completing the square for \(y\):
\[ y^2 + 12y = (y + 6)^2 - 36 \]

4. Substitute back into the equation:
\[ (x + \frac{3}{2})^2 - \frac{9}{4} + (y + 6)^2 - 36 = -2 \]

5. Simplifying:
\[ (x + \frac{3}{2})^2 + (y + 6)^2 = \frac{9}{4} + 36 - 2 \]
\[ (x + \frac{3}{2})^2 + (y + 6)^2 = \frac{9}{4} + \frac{144}{4} - \frac{8}{4} \]
\[ (x + \frac{3}{2})^2 + (y + 6)^2 = \frac{145}{4} \]

Thus, the center of the circle is \((- \frac{3}{2}, -6)\) and the radius is \(\sqrt{\frac{145}{4}} = \frac{\sqrt{145}}{2}\).

### Step 2: Find the midpoint of diagonal AC
Given the coordinates of points A and B:
- \(A(3, -2)\)
- \(B(-2, 0)\)

The midpoint \(O\) of diagonal \(AC\) can be found using the midpoint formula:
\[ O = \left( \frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2} \right) \]
Substituting the coordinates of A and B:
\[ O = \left( \frac{3 + (-2)}{2}, \frac{-2 + 0}{2} \right) = \left( \frac{1}{2}, -1 \right) \]

### Step 3: Find the coordinates of C and D
Since \(O\) is the midpoint of both diagonals \(AC\) and \(BD\), we can find the coordinates of \(C\) and \(D\) using the properties of the rectangle.

Let the coordinates of \(C\) be \((x_1, y_1)\) and the coordinates of \(D\) be \((x_2, y_2)\). Since \(O\) is the midpoint:
\[ O = \left( \frac{3 + x_1}{2}, \frac{-2 + y_1}{2} \right) \]
Setting this equal to the coordinates of \(O\):
1. From the x-coordinates:
\[ \frac{3 + x_1}{2} = \frac{1}{2} \implies 3 + x_1 = 1 \implies x_1 = 1 - 3 = -2 \]

2. From the y-coordinates:
\[ \frac{-2 + y_1}{2} = -1 \implies -2 + y_1 = -2 \implies y_1 = 0 \]
Thus, \(C(-2, 0)\).

For point \(D\):
\[ O = \left( \frac{-2 + x_2}{2}, \frac{0 + y_2}{2} \right) \]
Setting this equal to the coordinates of \(O\):
1. From the x-coordinates:
\[ \frac{-2 + x_2}{2} = \frac{1}{2} \implies -2 + x_2 = 1 \implies x_2 = 3 \]

2. From the y-coordinates:
\[ \frac{0 + y_2}{2} = -1 \implies y_2 = -2 \]
Thus, \(D(3, -2)\).

### Final Coordinates
The coordinates of the other two vertices of the rectangle are:
- \(C(-6, -10)\)
- \(D(-1, -12)\)