Find the equation of the circle the end point of whose diameter are (2,-3) and (2,4). Find its centre and radius.

Answer

To find the equation of the circle whose diameter endpoints are (2, -3) and (2, 4), we will 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 diameter endpoints. The midpoint (H, K) can be calculated using the formula:

\[ H = \frac{x_1 + x_2}{2}, \quad K = \frac{y_1 + y_2}{2} \]

Given points:
- \( (x_1, y_1) = (2, -3) \)
- \( (x_2, y_2) = (2, 4) \)

Calculating H and K:

\[ H = \frac{2 + 2}{2} = \frac{4}{2} = 2 \]
\[ K = \frac{-3 + 4}{2} = \frac{1}{2} \]

Thus, the center of the circle is \( (2, \frac{1}{2}) \).

### Step 2: Find the radius of the circle
The radius can be found by calculating half the length of the diameter. First, we need to find the length of the diameter using the distance formula:

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

Substituting the values:

\[ AB = \sqrt{(2 - 2)^2 + (4 - (-3))^2} = \sqrt{0 + (4 + 3)^2} = \sqrt{7^2} = 7 \]

Now, the radius \( r \) is half of the diameter:

\[ r = \frac{AB}{2} = \frac{7}{2} = 3.5 \]

### Step 3: Write the equation of the circle
The standard form of the equation of a circle is given by:

\[ (x - H)^2 + (y - K)^2 = r^2 \]

Substituting the values of H, K, and r:

\[ (x - 2)^2 + \left(y - \frac{1}{2}\right)^2 = (3.5)^2 \]

Calculating \( (3.5)^2 \):

\[ (3.5)^2 = 12.25 \]

Thus, the equation becomes:

\[ (x - 2)^2 + \left(y - \frac{1}{2}\right)^2 = 12.25 \]

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

\[ (x - 2)^2 = x^2 - 4x + 4 \]
\[ \left(y - \frac{1}{2}\right)^2 = y^2 - y + \frac{1}{4} \]

Combining these:

\[ x^2 - 4x + 4 + y^2 - y + \frac{1}{4} = 12.25 \]

Now, combine all terms:

\[ x^2 + y^2 - 4x - y + 4 + \frac{1}{4} = 12.25 \]

Convert \( 12.25 \) into a fraction:

\[ 12.25 = \frac{49}{4} \]

Now, we can write:

\[ x^2 + y^2 - 4x - y + 4 + \frac{1}{4} = \frac{49}{4} \]

To simplify, convert \( 4 \) into a fraction:

\[ 4 = \frac{16}{4} \]

Thus, we have:

\[ x^2 + y^2 - 4x - y + \frac{16}{4} + \frac{1}{4} = \frac{49}{4} \]

Combining gives:

\[ x^2 + y^2 - 4x - y + \frac{17}{4} = \frac{49}{4} \]

Subtract \( \frac{17}{4} \) from both sides:

\[ x^2 + y^2 - 4x - y = \frac{49}{4} - \frac{17}{4} = \frac{32}{4} = 8 \]

### Final Equation
Thus, the final equation of the circle is:

\[ x^2 + y^2 - 4x - y - 8 = 0 \]

### Summary
- Center: \( (2, \frac{1}{2}) \)
- Radius: \( 3.5 \)
- Equation: \( x^2 + y^2 - 4x - y - 8 = 0 \)