If a circle passes through the point `(a, b)` and cuts the circle `x^2 + y^2 = 4` orthogonally, then the locus of its centre is (a)`2ax+2by-(a^(2)+b^(2)+4)=0` (b) `2ax+2by-(a^(2)-b^(2)+k^(2))=0` (c) `x^(2)+y^(2)-3ax-4by+(a^(2)+b^(2)-k^(2))=0` (d) `x^(2)+y^(2)-2ax-3by+(a^(2)-b^(2)-k^(2))=0`

Options

A. `2ax+2by-(a^(2)+b^(2)+k^(2))=0`
B. `2ax+2by-(a^(2)-b^(2)+k^(2))=0`
C. `x^(2)+y^(2)-3ax-4by+(a^(2)+b^(2)-k^(2))=0`
D. `x^(2)+y^(2)-2ax-3by+(a^(2)-b^(2)-k^(2))=0`

Explanation

To find the locus of the center of a circle that passes through the point \((a, b)\) and cuts the circle \(x^2 + y^2 = 4\) orthogonally, we can follow these steps:

### Step 1: Understand the given circles
The first circle is given by the equation \(x^2 + y^2 = 4\). This can be rewritten in standard form as:
\[ (x - 0)^2 + (y - 0)^2 = 2^2 \]
This indicates that the center of this circle, \(C_2\), is at \((0, 0)\) and its radius \(R_2\) is \(2\).

### Step 2: Define the second circle
Let the center of the second circle be \((h, k)\) and its radius be \(R_1\). The equation of this circle can be written as:
\[ (x - h)^2 + (y - k)^2 = R_1^2 \]
Since this circle passes through the point \((a, b)\), we can substitute this point into the equation:
\[ (a - h)^2 + (b - k)^2 = R_1^2 \tag{1} \]

### Step 3: Apply the orthogonality condition
For two circles to cut each other orthogonally, the following condition must hold:
\[ R_1^2 + R_2^2 = d^2 \]
where \(d\) is the distance between the centers of the circles. The distance \(d\) between the centers \((h, k)\) and \((0, 0)\) is given by:
\[ d = \sqrt{h^2 + k^2} \]
Thus, the orthogonality condition becomes:
\[ R_1^2 + 2^2 = h^2 + k^2 \tag{2} \]

### Step 4: Substitute \(R_1^2\) from equation (1) into (2)
From equation (1), we have:
\[ R_1^2 = (a - h)^2 + (b - k)^2 \]
Substituting this into equation (2):
\[ (a - h)^2 + (b - k)^2 + 4 = h^2 + k^2 \]

### Step 5: Expand and simplify
Expanding the left side:
\[ (a^2 - 2ah + h^2) + (b^2 - 2bk + k^2) + 4 = h^2 + k^2 \]
This simplifies to:
\[ a^2 + b^2 - 2ah - 2bk + 4 = 0 \]
Rearranging gives:
\[ 2ah + 2bk = a^2 + b^2 + 4 \]

### Step 6: Rearranging to find the locus
Rearranging the equation gives us:
\[ 2ax + 2by - (a^2 + b^2 + 4) = 0 \]
This is the equation of the locus of the center of the circle.

### Final Answer
Thus, the locus of the center of the circle is:
\[ \boxed{2ax + 2by - (a^2 + b^2 + 4) = 0} \]