Let $A, B, C$ be three points on a circle of radius $1$ such that $\angle ACB = \frac{\pi}{4}$. Then,the length of the side $AB$ 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:

Consider a circle $C$ which touches the $y$-axis at $(0,6)$ and cuts off an intercept $6 \sqrt{5}$ on the $x$-axis. Then the radius of the circle $C$ is equal to:

The length of the shortest path that begins at the point $(2, 5)$,touches the $x-$axis,and then ends at a point on the circle $x^2 + y^2 + 12x - 20y + 120 = 0$.

The minimum distance and maximum distance of the point $P(2,-7)$ from the circle $x^2+y^2-14x-10y-151=0$ are respectively . . . . . . units.

In a circle with center $O$,suppose $A, P, B$ are three points on its circumference such that $P$ is the mid-point of the minor arc $AB$. Suppose when $\angle AOB = \theta$,$\frac{\text{area}(\triangle AOB)}{\text{area}(\triangle APB)} = \sqrt{5} + 2$. If $\angle AOB$ is doubled to $2\theta$,then the ratio $\frac{\text{area}(\triangle AOB)}{\text{area}(\triangle APB)}$ is