The locus of the centre of the circle passing through the intersection of the circles `x^2+y^2= 1` and `x^2 + y^2-2x+y=0` is

Options

A. `x+2y=0`
B. `2x-y=1`
C. a circle
D. a pair of lines

Explanation

To find the locus of the center of the circle passing through the intersection of the circles given by the equations \(x^2 + y^2 = 1\) and \(x^2 + y^2 - 2x + y = 0\), we can follow these steps:

### Step 1: Write the equations of the circles
The first circle is given by:
\[ x^2 + y^2 = 1 \]
The second circle can be rewritten by completing the square:
\[ x^2 - 2x + y^2 + y = 0 \implies (x-1)^2 + (y + \frac{1}{2})^2 = \frac{5}{4} \]
This shows that the second circle is centered at \((1, -\frac{1}{2})\) with a radius of \(\frac{\sqrt{5}}{2}\).

### Step 2: Find the equation of the family of circles passing through the intersection
The equation of the family of circles passing through the intersection of the two given circles can be expressed as:
\[ x^2 + y^2 - 2x + y + \lambda(2x - y - 1) = 0 \]
where \(\lambda\) is a parameter.

### Step 3: Rearranging the equation
Expanding and rearranging gives:
\[ x^2 + y^2 + (2 - 2\lambda)x + (1 - \lambda)y - \lambda = 0 \]

### Step 4: Identify the center of the circle
The general equation of a circle can be written as:
\[ x^2 + y^2 + 2gx + 2fy + c = 0 \]
From our rearranged equation, we can identify:
- \(g = 1 - \lambda\)
- \(f = \frac{1 - \lambda}{2}\)
- \(c = -\lambda\)

### Step 5: Find the coordinates of the center
The center of the circle \((h, k)\) can be expressed in terms of \(g\) and \(f\):
\[ h = -2g = -2(1 - \lambda) = -2 + 2\lambda \]
\[ k = -2f = -2\left(\frac{1 - \lambda}{2}\right) = -1 + \lambda \]

### Step 6: Express the locus in terms of \(h\) and \(k\)
From the equations for \(h\) and \(k\):
\[ \lambda = k + 1 \]
Substituting \(\lambda\) back into the equation for \(h\):
\[ h = -2 + 2(k + 1) = -2 + 2k + 2 = 2k \]

### Step 7: Form the locus equation
Thus, we have:
\[ h = 2k \implies h - 2k = 0 \]
If we replace \(h\) with \(x\) and \(k\) with \(y\), we get:
\[ x - 2y = 0 \implies x + 2y = 0 \]

### Final Answer
The locus of the center of the circle passing through the intersection of the given circles is:
\[ \boxed{x + 2y = 0} \]