Fill up in each of the following
(c) the equation of circle whose end point of of one of its diameter are (-2, 3) and (0,-1) is

Answer

To find the equation of the circle whose endpoints of one of its diameters are given as (-2, 3) and (0, -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 formula for the midpoint \( M(x, y) \) between two points \( (x_1, y_1) \) and \( (x_2, y_2) \) is given by:

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

For the points (-2, 3) and (0, -1):
- \( x_1 = -2 \), \( y_1 = 3 \)
- \( x_2 = 0 \), \( y_2 = -1 \)

Calculating the midpoint:

\[ M = \left( \frac{-2 + 0}{2}, \frac{3 + (-1)}{2} \right) = \left( \frac{-2}{2}, \frac{2}{2} \right) = (-1, 1) \]

### 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 \( r \) between two points \( (x_1, y_1) \) and \( (x_2, y_2) \) is given by:

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

Using the center (-1, 1) and one endpoint (0, -1):

\[ r = \sqrt{(0 - (-1))^2 + (-1 - 1)^2} = \sqrt{(0 + 1)^2 + (-2)^2} = \sqrt{1 + 4} = \sqrt{5} \]

### Step 3: 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 \( r = \sqrt{5} \):

\[ (x - (-1))^2 + (y - 1)^2 = (\sqrt{5})^2 \]

This simplifies to:

\[ (x + 1)^2 + (y - 1)^2 = 5 \]

### Step 4: Expand the equation (optional)
If we want to expand the equation:

\[ (x + 1)^2 + (y - 1)^2 = 5 \]

Expanding gives:

\[ x^2 + 2x + 1 + y^2 - 2y + 1 = 5 \]

Combining like terms:

\[ x^2 + y^2 + 2x - 2y + 2 - 5 = 0 \]

This simplifies to:

\[ x^2 + y^2 + 2x - 2y - 3 = 0 \]

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

\[ x^2 + y^2 + 2x - 2y - 3 = 0 \]