The circle with centre (1,1) and radius 1 is

Options

A. `x^(2)+y^(2)-2x-2y+1=0`
B. `x^(2)+y^(2)+2x+2y+1=0`
C. `x^(2)+y^(2)_x-y+1=0`
D. `x^(2)+y^(2)+x+y+1=0`

Answer

To find the equation of the circle with center (1, 1) and radius 1, we can use the standard form of the equation of a circle. Here’s a step-by-step solution:

### Step 1: Identify the center and radius
The center of the circle is given as (1, 1), which means:
- \( h = 1 \)
- \( k = 1 \)

The radius \( r \) is given as 1.

### Step 2: Write the standard form of the circle's equation
The standard form of the equation of a circle with center (h, k) and radius r is:
\[ (x - h)^2 + (y - k)^2 = r^2 \]

### Step 3: Substitute the values into the equation
Substituting the values of \( h \), \( k \), and \( r \) into the equation:
\[ (x - 1)^2 + (y - 1)^2 = 1^2 \]
This simplifies to:
\[ (x - 1)^2 + (y - 1)^2 = 1 \]

### Step 4: Expand the equation
Now, we will expand both squares:
\[ (x - 1)^2 = x^2 - 2x + 1 \]
\[ (y - 1)^2 = y^2 - 2y + 1 \]
So, adding these together gives:
\[ x^2 - 2x + 1 + y^2 - 2y + 1 = 1 \]

### Step 5: Combine like terms
Combine the constant terms:
\[ x^2 + y^2 - 2x - 2y + 2 = 1 \]

### Step 6: Rearrange the equation
Now, we will move 1 to the left side:
\[ x^2 + y^2 - 2x - 2y + 2 - 1 = 0 \]
This simplifies to:
\[ x^2 + y^2 - 2x - 2y + 1 = 0 \]

### Final Answer
Thus, the equation of the circle is:
\[ x^2 + y^2 - 2x - 2y + 1 = 0 \]