If (0,1) and (1,1) are ends points of a diameter of a circle then its equation is

Options

A. `x^2+y^2-x-2y+1=0`
B. `x^2+y^2+x-2y+1=0`
C. `x^2+y^2-x-2y-1=0`
D. none of these

Answer

To find the equation of the circle with endpoints of a diameter at the points (0, 1) and (1, 1), we can follow these steps:

### Step 1: Find the center of the circle
The center of the circle can be found using the midpoint formula. If (x1, y1) and (x2, y2) are the endpoints of the diameter, the center (C) is given by:

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

Substituting the given points (0, 1) and (1, 1):

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

### Step 2: Find the radius of the circle
The radius (r) can be calculated as the distance from the center to one of the endpoints. We can use the distance formula:

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

Using the center (1/2, 1) and one of the endpoints (0, 1):

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

### Step 3: Write the equation of the circle
The standard form of the equation of a circle with center (a, b) and radius r is given by:

\[ (x - a)^2 + (y - b)^2 = r^2 \]

Substituting the center (1/2, 1) and radius (1/2):

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

This simplifies to:

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

### Step 4: Expand the equation
Now, we will expand the equation:

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

Combining these, we have:

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

Subtracting \(\frac{1}{4}\) from both sides gives:

\[ x^2 + y^2 - x - 2y + 1 = 0 \]

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

\[ x^2 + y^2 - x - 2y + 1 = 0 \]