Show that the line `(x-2)costheta+(y-2)sintheta=1` touches a circle for all values of `theta` .Find the circle.

Answer

To show that the line \((x-2)\cos\theta + (y-2)\sin\theta = 1\) touches a circle for all values of \(\theta\) and to find the equation of that circle, we can follow these steps:

### Step 1: Rewrite the given line equation
The given line equation is:
\[ (x-2)\cos\theta + (y-2)\sin\theta = 1 \]

### Step 2: Rearranging the equation
We can rearrange this equation to express it in a standard form:
\[ x\cos\theta + y\sin\theta - 2\cos\theta - 2\sin\theta = 1 \]
This can be written as:
\[ x\cos\theta + y\sin\theta = 1 + 2\cos\theta + 2\sin\theta \]

### Step 3: Use the identity \(\cos^2\theta + \sin^2\theta = 1\)
We know that \(\cos^2\theta + \sin^2\theta = 1\). We can use this identity to relate the terms involving \(\theta\).

### Step 4: Set up the equations
From the rearranged equation, we can set:
\[ x - 2\cos\theta = \cos^2\theta \quad \text{(1)} \]
\[ y - 2\sin\theta = \sin^2\theta \quad \text{(2)} \]

### Step 5: Square and add the equations
Now, we square both equations (1) and (2) and add them:
\[ (x - 2\cos\theta)^2 + (y - 2\sin\theta)^2 = \cos^4\theta + \sin^4\theta \]

### Step 6: Simplify the left-hand side
Expanding the left-hand side:
\[ (x^2 - 4x\cos\theta + 4\cos^2\theta) + (y^2 - 4y\sin\theta + 4\sin^2\theta) \]
This gives us:
\[ x^2 + y^2 - 4x\cos\theta - 4y\sin\theta + 4(\cos^2\theta + \sin^2\theta) \]
Using \(\cos^2\theta + \sin^2\theta = 1\), we can simplify this to:
\[ x^2 + y^2 - 4x\cos\theta - 4y\sin\theta + 4 \]

### Step 7: Set the right-hand side
The right-hand side can be simplified using the identity for \(\cos^4\theta + \sin^4\theta\):
\[ \cos^4\theta + \sin^4\theta = (\cos^2\theta + \sin^2\theta)^2 - 2\cos^2\theta\sin^2\theta = 1 - 2\cos^2\theta\sin^2\theta \]

### Step 8: Combine and find the circle
Now we have:
\[ x^2 + y^2 - 4x\cos\theta - 4y\sin\theta + 4 = 1 - 2\cos^2\theta\sin^2\theta \]
This implies that the left-hand side represents a circle centered at \((2\cos\theta, 2\sin\theta)\) with a radius that depends on \(\theta\).

### Final Circle Equation
The equation of the circle can be expressed as:
\[ (x - 2\cos\theta)^2 + (y - 2\sin\theta)^2 = r^2 \]
where \(r^2 = 1 - 2\cos^2\theta\sin^2\theta\).

### Conclusion
Thus, we have shown that the line touches a circle for all values of \(\theta\) and the center of the circle is \((2\cos\theta, 2\sin\theta)\).

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