The locus of the centre of the circle`(xcos alpha + y sin alpha - a)^2 + (x sin alpha - y cos alpha - b)^2= k^2` if `alpha` varies, is

Options

A. `x^(2)-y^(2)=a^(2)+b^(2)`
B. `x^(2)-y^(2)=a^(2)b^(2)`
C. `x^(2)+y^(2)=a^(2)+b^(2)`
D. `x^(2)+y^(2)=a^(2)b^(2)`

Explanation

To find the locus of the center of the circle given by the equation

\[ (x \cos \alpha + y \sin \alpha - a)^2 + (x \sin \alpha - y \cos \alpha - b)^2 = k^2 \]

where \(\alpha\) varies, we will follow these steps:

### Step 1: Identify the center of the circle
The center of the circle can be expressed in terms of \(\alpha\). We can rewrite the equation in a more manageable form by recognizing that the center \((h, k)\) can be represented as:

\[ h = a \cos \alpha + b \sin \alpha \]
\[ k = a \sin \alpha - b \cos \alpha \]

### Step 2: Square the expressions for \(h\) and \(k\)
Now, we will square both expressions for \(h\) and \(k\) and add them together:

\[ h^2 + k^2 = (a \cos \alpha + b \sin \alpha)^2 + (a \sin \alpha - b \cos \alpha)^2 \]

### Step 3: Expand the squared terms
Expanding both squared terms:

\[ h^2 = a^2 \cos^2 \alpha + 2ab \cos \alpha \sin \alpha + b^2 \sin^2 \alpha \]
\[ k^2 = a^2 \sin^2 \alpha - 2ab \sin \alpha \cos \alpha + b^2 \cos^2 \alpha \]

### Step 4: Combine the expressions
Now, we combine the two expansions:

\[ h^2 + k^2 = (a^2 \cos^2 \alpha + b^2 \sin^2 \alpha + 2ab \cos \alpha \sin \alpha) + (a^2 \sin^2 \alpha + b^2 \cos^2 \alpha - 2ab \sin \alpha \cos \alpha) \]

### Step 5: Simplify the expression
Notice that the terms \(2ab \cos \alpha \sin \alpha\) and \(-2ab \sin \alpha \cos \alpha\) cancel each other out. Thus, we have:

\[ h^2 + k^2 = a^2 (\cos^2 \alpha + \sin^2 \alpha) + b^2 (\sin^2 \alpha + \cos^2 \alpha) \]

Using the Pythagorean identity \(\cos^2 \alpha + \sin^2 \alpha = 1\):

\[ h^2 + k^2 = a^2 + b^2 \]

### Step 6: Write the locus equation
Now, we can express the locus of the center of the circle as:

\[ x^2 + y^2 = a^2 + b^2 \]

### Conclusion
Thus, the locus of the center of the circle is given by:

\[ x^2 + y^2 = a^2 + b^2 \]