Find the equation of a circle whose diameter has end points (4, 3) and (-2, 1).

Options

A. a)`x^(2)+y^(2)-2x+4y=5`
B. b)`x^(2)+y^(2)-6x+2y=3`
C. c)`x^(2)+y^(2)-2x-4y=5`
D. d)`x^(2)+y^(2)-2x-4y=3`

Answer

To find the equation of a circle whose diameter has endpoints (4, 3) and (-2, 1), we can follow these steps:

### Step 1: Identify the endpoints of the diameter
The endpoints of the diameter are given as:
- Point 1: \( (x_1, y_1) = (4, 3) \)
- Point 2: \( (x_2, y_2) = (-2, 1) \)

### Step 2: Calculate the center of the circle
The center of the circle is the midpoint of the diameter. The midpoint \( (h, k) \) can be calculated using the formula:
\[ h = \frac{x_1 + x_2}{2}, \quad k = \frac{y_1 + y_2}{2} \]
Substituting the values:
\[ h = \frac{4 + (-2)}{2} = \frac{2}{2} = 1 \]
\[ k = \frac{3 + 1}{2} = \frac{4}{2} = 2 \]
Thus, the center of the circle is \( (1, 2) \).

### Step 3: Calculate the radius of the circle
The radius \( r \) of the circle is half the distance between the endpoints of the diameter. We first calculate the distance \( d \) between the two points using the distance formula:
\[ d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \]
Substituting the values:
\[ d = \sqrt{((-2) - 4)^2 + (1 - 3)^2} = \sqrt{(-6)^2 + (-2)^2} = \sqrt{36 + 4} = \sqrt{40} = 2\sqrt{10} \]
Thus, the radius \( r \) is:
\[ r = \frac{d}{2} = \frac{2\sqrt{10}}{2} = \sqrt{10} \]

### 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 \( h = 1 \), \( k = 2 \), and \( r = \sqrt{10} \):
\[ (x - 1)^2 + (y - 2)^2 = (\sqrt{10})^2 \]
This simplifies to:
\[ (x - 1)^2 + (y - 2)^2 = 10 \]

### Final Answer
The equation of the circle is:
\[ (x - 1)^2 + (y - 2)^2 = 10 \]

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