If (4,3) and (-12,-1) are the end points of a diameter of a circle then the equation of the circle is

Options

A. `x^2+y^2-8x-2y-51=0`
B. `x^2+y^2+8x-2y-51=0`
C. `x^2+y^2+8x+2y-51=0`
D. none of these

Answer

To find the equation of the circle with endpoints of the diameter at points (4, 3) and (-12, -1), we can follow these steps:

### Step 1: Find the center of the circle
The center of the circle can be found by calculating the midpoint of the endpoints of the diameter. The midpoint (M) of two points \((x_1, y_1)\) and \((x_2, y_2)\) is given by:

\[ M = \left( \frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2} \right) \]

Here, the points are \( (4, 3) \) and \( (-12, -1) \).

Calculating the midpoint:

\[ M = \left( \frac{4 + (-12)}{2}, \frac{3 + (-1)}{2} \right) = \left( \frac{4 - 12}{2}, \frac{3 - 1}{2} \right) = \left( \frac{-8}{2}, \frac{2}{2} \right) = (-4, 1) \]

### Step 2: Find the radius of the circle
The radius can be found by calculating the distance from the center to one of the endpoints. The distance \(d\) between two points \((x_1, y_1)\) and \((x_2, y_2)\) is given by:

\[ d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \]

Using the center \((-4, 1)\) and one endpoint \( (4, 3) \):

\[ r = \sqrt{(4 - (-4))^2 + (3 - 1)^2} = \sqrt{(4 + 4)^2 + (3 - 1)^2} = \sqrt{(8)^2 + (2)^2} = \sqrt{64 + 4} = \sqrt{68} \]

### Step 3: Write the equation of the circle
The general equation of a circle with center \((h, k)\) and radius \(r\) is:

\[ (x - h)^2 + (y - k)^2 = r^2 \]

Substituting \(h = -4\), \(k = 1\), and \(r^2 = 68\):

\[ (x + 4)^2 + (y - 1)^2 = 68 \]

### Step 4: Expand the equation
Expanding the equation:

\[ (x + 4)^2 + (y - 1)^2 = 68 \]

Expanding each term:

\[ (x^2 + 8x + 16) + (y^2 - 2y + 1) = 68 \]

Combining like terms:

\[ x^2 + y^2 + 8x - 2y + 17 = 68 \]

Subtracting 68 from both sides:

\[ x^2 + y^2 + 8x - 2y + 17 - 68 = 0 \]

Simplifying:

\[ x^2 + y^2 + 8x - 2y - 51 = 0 \]

### Final Answer
The equation of the circle is:

\[ x^2 + y^2 + 8x - 2y - 51 = 0 \]