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

Options

A. `x^(2)+y^(2)-3x-3y+1=0`
B. `2x+2y-1=0`
C. `x^(2)+y^(2)-2x-2y+1=0`
D. `2x+2y-3=0`

Answer

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

### Step 1: Define the circles
Let the first circle be \(C_1: x^2 + y^2 = 1\), which has its center at \(O_2(0, 0)\) and radius \(R_2 = 1\).

Let the second circle \(C_2\) have its center at \(O_1(x, y)\) and radius \(R_1\). Since it passes through the point (1, 1), we can express \(R_1\) as:
\[ R_1 = \sqrt{(x - 1)^2 + (y - 1)^2} \]

### Step 2: Use the orthogonality condition
For two circles to intersect orthogonally, the following condition must hold:
\[ R_1^2 + R_2^2 = O_1O_2^2 \]
Where \(O_1O_2\) is the distance between the centers of the two circles. We can calculate \(O_1O_2\) as:
\[ O_1O_2 = \sqrt{x^2 + y^2} \]
Thus,
\[ O_1O_2^2 = x^2 + y^2 \]

### Step 3: Substitute the values into the orthogonality condition
Now substituting the values of \(R_1^2\), \(R_2^2\), and \(O_1O_2^2\) into the orthogonality condition:
\[ R_1^2 + R_2^2 = O_1O_2^2 \]
This gives us:
\[ (x - 1)^2 + (y - 1)^2 + 1 = x^2 + y^2 \]

### Step 4: Expand and simplify the equation
Expanding \(R_1^2\):
\[ (x - 1)^2 = x^2 - 2x + 1 \]
\[ (y - 1)^2 = y^2 - 2y + 1 \]
Thus, we have:
\[ (x^2 - 2x + 1) + (y^2 - 2y + 1) + 1 = x^2 + y^2 \]
Combining like terms:
\[ x^2 - 2x + 1 + y^2 - 2y + 1 + 1 = x^2 + y^2 \]
This simplifies to:
\[ -2x - 2y + 3 = 0 \]

### Step 5: Rearranging the equation
Rearranging gives us:
\[ 2x + 2y - 3 = 0 \]
Dividing through by 2:
\[ x + y = \frac{3}{2} \]

### Conclusion
The locus of the center of the circle is given by the line:
\[ x + y = \frac{3}{2} \]