If a circle passes through the point (a,b) and cuts the circle `x^(2)+y^(2)=4` orthogonally, then locus of its centre is

Options

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

Answer

To solve the problem, we need 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.

### Step-by-Step Solution:

1. **Equation of the Circle**:
Let the equation of the circle that passes through the point (a, b) be given in the standard form:
\[ x^2 + y^2 + 2gx + 2fy + c = 0 \]
Here, the center of the circle is \((-g, -f)\).

2. **Condition for Passing Through (a, b)**:
Since the circle passes through the point (a, b), we substitute \(x = a\) and \(y = b\) into the equation:
\[ a^2 + b^2 + 2ga + 2fb + c = 0 \tag{1} \]

3. **Condition for Orthogonality**:
The circle \(x^2 + y^2 = 4\) can be rewritten as:
\[ x^2 + y^2 + 0x + 0y - 4 = 0 \]
For the two circles to cut orthogonally, the following condition must hold:
\[ 2g \cdot 0 + 2f \cdot 0 = c + (-4) \]
This simplifies to:
\[ c = 4 \tag{2} \]

4. **Substituting c in Equation (1)**:
Now, substitute \(c = 4\) into equation (1):
\[ a^2 + b^2 + 2ga + 2fb + 4 = 0 \]
Rearranging gives:
\[ a^2 + b^2 + 2ga + 2fb = -4 \tag{3} \]

5. **Expressing in Terms of g and f**:
We can express \(g\) and \(f\) in terms of \(x\) and \(y\) (the coordinates of the center):
\[ g = -x, \quad f = -y \]
Substituting these into equation (3):
\[ a^2 + b^2 - 2ax - 2by = -4 \]
Rearranging gives:
\[ 2ax + 2by + a^2 + b^2 + 4 = 0 \tag{4} \]

6. **Final Form of the Locus**:
The equation (4) represents the locus of the center of the circle. We can rewrite it as:
\[ 2ax + 2by + (a^2 + b^2 + 4) = 0 \]
This is the required locus of the center of the circle.

### Conclusion:
The locus of the center of the circle that passes through the point (a, b) and cuts the circle \(x^2 + y^2 = 4\) orthogonally is given by:
\[ 2ax + 2by + (a^2 + b^2 + 4) = 0 \]