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

Options

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

Answer

To solve the problem step by step, we will find the locus of the center of the circle that passes through the point (1, 2) and cuts the circle \(x^2 + y^2 = 4\) orthogonally.

### Step 1: Understand the given circles
The first circle is given by the equation:
\[ x^2 + y^2 = 4 \]
This circle has:
- Center: \(C_1(0, 0)\)
- Radius: \(r_1 = 2\)

Let the second circle be represented by the equation:
\[ (x - h)^2 + (y - k)^2 = r^2 \]
where \((h, k)\) is the center of the second circle and \(r\) is its radius.

### Step 2: Condition for orthogonality
Two circles cut each other orthogonally if the following condition holds:
\[ r_1^2 + r_2^2 = d^2 \]
where \(d\) is the distance between the centers of the two circles.

Here, we have:
- \(r_1^2 = 4\)
- \(r_2^2 = r^2\)
- The distance \(d\) between the centers \(C_1(0, 0)\) and \(C_2(h, k)\) is given by:
\[ d = \sqrt{h^2 + k^2} \]
Thus, the condition becomes:
\[ 4 + r^2 = h^2 + k^2 \tag{1} \]

### Step 3: Circle passing through the point (1, 2)
Since the second circle passes through the point (1, 2), we can substitute this point into the circle's equation:
\[ (1 - h)^2 + (2 - k)^2 = r^2 \]
Expanding this gives:
\[ (1 - h)^2 + (2 - k)^2 = r^2 \]
\[ 1 - 2h + h^2 + 4 - 4k + k^2 = r^2 \]
\[ h^2 + k^2 - 2h - 4k + 5 = r^2 \tag{2} \]

### Step 4: Substitute \(r^2\) from equation (1) into equation (2)
From equation (1), we have \(r^2 = h^2 + k^2 - 4\). Substituting this into equation (2):
\[ h^2 + k^2 - 2h - 4k + 5 = h^2 + k^2 - 4 \]
Now, simplify:
\[ -2h - 4k + 5 = -4 \]
\[ -2h - 4k + 9 = 0 \]
Dividing the entire equation by -1 gives:
\[ 2h + 4k - 9 = 0 \]
or
\[ h + 2k = \frac{9}{2} \tag{3} \]

### Step 5: Locus of the center
The equation \(h + 2k = \frac{9}{2}\) represents the locus of the center of the second circle. In standard form, we can write it as:
\[ h + 2k - \frac{9}{2} = 0 \]

### Final Answer
The locus of the center of the circle is:
\[ h + 2k - \frac{9}{2} = 0 \]