If a circle Passes through a point (1,2) and cut the circle `x^2+y^2 = 4` orthogonally,Then the locus of its centre is

Options

A. `x^(2)+y^(2)-3x-8y+1=0`
B. `x^(2)+y^(2)-2x-6y-7=0`
C. `2x+4y-9=0`
D. `2x+4y-1=0`

Explanation

To solve the problem, we need to find the locus of the center of a circle that passes through the point (1, 2) and intersects the circle defined by the equation \(x^2 + y^2 = 4\) orthogonally.

### Step-by-Step Solution:

1. **Assume the Equation of the Circle:**
Let the equation of the circle be:
\[ x^2 + y^2 + 2gx + 2fy + c = 0 \]
where \((-g, -f)\) is the center of the circle.

2. **Condition for Orthogonal Intersection:**
For two circles to intersect orthogonally, the following condition must hold:
\[ 2g_1g_2 + 2f_1f_2 = c_1 + c_2 \]
Here, we will compare our circle with the given circle \(x^2 + y^2 = 4\), which can be rewritten as:
\[ x^2 + y^2 + 0x + 0y - 4 = 0 \]
From this, we identify:
- \(g_2 = 0\)
- \(f_2 = 0\)
- \(c_2 = -4\)

3. **Identify Parameters from Our Circle:**
From our assumed circle, we have:
- \(g_1 = g\)
- \(f_1 = f\)
- \(c_1 = c\)

4. **Substituting into the Orthogonality Condition:**
Substituting the values into the orthogonality condition:
\[ 2g \cdot 0 + 2f \cdot 0 = c - 4 \]
This simplifies to:
\[ 0 = c - 4 \implies c = 4 \]

5. **Substituting \(c\) Back into the Circle Equation:**
Now substituting \(c = 4\) back into the equation of our circle:
\[ x^2 + y^2 + 2gx + 2fy + 4 = 0 \]

6. **Using the Point (1, 2):**
Since the circle passes through the point (1, 2), we substitute \(x = 1\) and \(y = 2\) into the circle equation:
\[ 1^2 + 2^2 + 2g(1) + 2f(2) + 4 = 0 \]
This simplifies to:
\[ 1 + 4 + 2g + 4f + 4 = 0 \implies 2g + 4f + 9 = 0 \]

7. **Expressing in Terms of \(g\) and \(f\):**
Rearranging gives:
\[ 2g + 4f = -9 \]

8. **Finding the Locus of the Center:**
Since the center of the circle is \((-g, -f)\), we can express \(g\) and \(f\) in terms of the coordinates of the center \((h, k)\):
\[ g = -h, \quad f = -k \]
Substituting these into the equation:
\[ 2(-h) + 4(-k) = -9 \implies -2h - 4k = -9 \implies 2h + 4k = 9 \]

9. **Final Locus Equation:**
Thus, the locus of the center of the circle is:
\[ 2x + 4y - 9 = 0 \]

### Conclusion:
The locus of the center of the circle that passes through the point (1, 2) and intersects the circle \(x^2 + y^2 = 4\) orthogonally is given by the equation:
\[ 2x + 4y - 9 = 0 \]