M,N,O,P are circle `x^2+y^2=1 , x^2+y^2-2x=0 , x^2+y^2-2y=0 , x^2+y^2-2x-2y=0` Centers of thses circles are joined then shape formed is

Options

A. rhombus
B. rectangle
C. square
D. Parallelogram

Answer

To solve the problem, we need to find the centers of the given circles and then determine the shape formed by connecting these centers.

### Step-by-Step Solution:

1. **Identify the equations of the circles**:
- Circle M: \( x^2 + y^2 = 1 \)
- Circle N: \( x^2 + y^2 - 2x = 0 \)
- Circle O: \( x^2 + y^2 - 2y = 0 \)
- Circle P: \( x^2 + y^2 - 2x - 2y = 0 \)

2. **Find the centers of the circles**:
- For Circle M: The equation is \( x^2 + y^2 = 1 \). The center is at \( (0, 0) \).
- For Circle N: Rewrite as \( x^2 + y^2 = 2x \) or \( x^2 - 2x + y^2 = 0 \). Completing the square gives \( (x-1)^2 + y^2 = 1 \). The center is at \( (1, 0) \).
- For Circle O: Rewrite as \( x^2 + y^2 = 2y \) or \( x^2 + (y-1)^2 = 1 \). The center is at \( (0, 1) \).
- For Circle P: Rewrite as \( x^2 + y^2 = 2x + 2y \) or \( (x-1)^2 + (y-1)^2 = 2 \). The center is at \( (1, 1) \).

3. **List the centers**:
- Center M: \( (0, 0) \)
- Center N: \( (1, 0) \)
- Center O: \( (0, 1) \)
- Center P: \( (1, 1) \)

4. **Plot the points**:
- Plot the points M, N, O, and P on the Cartesian plane:
- M: \( (0, 0) \)
- N: \( (1, 0) \)
- O: \( (0, 1) \)
- P: \( (1, 1) \)

5. **Connect the points**:
- Connect the points M, N, O, and P in order.

6. **Determine the shape**:
- The points form a square since:
- All sides (MN, NO, OP, PM) are equal (length = 1).
- All angles are right angles (90 degrees).

### Conclusion:
The shape formed by joining the centers of the circles M, N, O, and P is a **square**.