The point diametrically opposite to the point $P(1, 0)$ on the circle $x^2 + y^2 + 2x + 4y - 3 = 0$ is

Find the equation of a circle that touches the $Y$-axis at a distance of $4$ units from the origin and cuts an intercept of $6$ units on the $X$-axis.

Suppose we have two circles of radius $2$ each in the plane such that the distance between their centers is $2 \sqrt{3}$. The area of the region common to both circles lies between

The area (in square units) of the circle which touches the lines $4x + 3y = 15$ and $4x + 3y = 5$ is

For the 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 circle $N$,the centre of circle $N$ is joined with the centre of circle $O$,the 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:

Chords $AB$ and $CD$ of a circle intersect at a right angle at point $P$. If the lengths of $AP$, $PB$, $CP$, and $PD$ are $2$, $6$, $3$, and $4$ units respectively, then the radius of the circle is: