Question

For all four circles $M,N,O$ and $P$, the following four equations are given:

Circle $M:x^2+y^2=1$

Circle $N:x^2+y^2-2x=0$

Circle $O:x^2+y^2-2x-2y+1=0$

Circle $P:x^2+y^2-2y=0$

If the centre of circle $M$ is joined with the centre of the circle $N$, further centre of circle $N$ is joined with the centre of the circle $O$, centre of circle $O$ is joined with the centre of circle $P$ and lastly, the centre of circle $P$ is joined with the centre of circle $M$, then these lines form the sides of $a$:

• A. Rectangle
• B. Square
• C. Parallelogram
• D. Rhombus

Answer 1

The correct option is B

Square

Step 1: Find the centre of all the given circles.

General form of the circle is$x^2+y^2+2gx+2fy+c=0$

then centre will$(-g,-f)$.

Circle $M:x^2+y^2=1$

it's centre clearly, passing through origin.

Centre of $M:(0,0)$

Circle $N:x^2+y^2-2x=0$

Centre of $N:(1,0)$

Circle $O:x^2+y^2-2x-2y+1=0$

Centre of $O:(1,1)$

Circle $P:x^2+y^2-2y=0$

Centre of $P:(0,1)$

Step 2: Join the all centres of the given circles and check the shape.

By joining all the centres, we get $MNOP$ a quadrilateral.

By using distance formula we will find that $MNOP$ will be forming a square.

Hence, option $\left(B\right)$is the correct option.