The line `(x-1) cos theta + (y-1) sin theta =1,` for all values of `theta` touches the circle

Options

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

Answer

To solve the problem, we need to show that the line given by \((x-1) \cos \theta + (y-1) \sin \theta = 1\) touches a circle. Let's derive the equation of the circle step by step.

### Step 1: Rewrite the line equation
The line equation is given as:
\[ (x-1) \cos \theta + (y-1) \sin \theta = 1 \]
This can be rearranged to express it in a standard form.

### Step 2: Substitute \(1\) with \(\sin^2 \theta + \cos^2 \theta\)
Using the Pythagorean identity \(\sin^2 \theta + \cos^2 \theta = 1\), we can rewrite the equation:
\[ (x-1) \cos \theta + (y-1) \sin \theta = \sin^2 \theta + \cos^2 \theta \]

### Step 3: Expand the equation
Expanding both sides gives:
\[ (x-1) \cos \theta + (y-1) \sin \theta = \cos^2 \theta + \sin^2 \theta \]
This simplifies to:
\[ (x-1) \cos \theta + (y-1) \sin \theta = 1 \]

### Step 4: Compare coefficients
From the equation, we can compare coefficients:
- The coefficient of \(\cos \theta\) gives us \(x - 1 = \cos \theta\)
- The coefficient of \(\sin \theta\) gives us \(y - 1 = \sin \theta\)

### Step 5: Express \(x\) and \(y\)
From the above comparisons, we can express \(x\) and \(y\) as:
\[ x = 1 + \cos \theta \]
\[ y = 1 + \sin \theta \]

### Step 6: Form the equation of the circle
Now, we can write the equation of the circle using the values of \(x\) and \(y\):
\[ (x - 1)^2 + (y - 1)^2 = \cos^2 \theta + \sin^2 \theta \]
Since \(\cos^2 \theta + \sin^2 \theta = 1\), we have:
\[ (x - 1)^2 + (y - 1)^2 = 1 \]

### Step 7: Expand the circle equation
Expanding the equation gives:
\[ (x - 1)^2 + (y - 1)^2 = 1 \]
\[ x^2 - 2x + 1 + y^2 - 2y + 1 = 1 \]
Combining like terms results in:
\[ x^2 + y^2 - 2x - 2y + 2 = 1 \]
Subtracting 1 from both sides:
\[ x^2 + y^2 - 2x - 2y + 1 = 0 \]

### Final Equation
The final equation of the circle is:
\[ x^2 + y^2 - 2x - 2y + 1 = 0 \]

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