Let the centre of the circle be in the first quadrant and lie on the line $2x - y = 4$. Let the area of an equilateral triangle inscribed in the circle be $27\sqrt{3}$. Then the square of the length of the chord of the circle on the line $x = 1$ is . . . . . .

The mid-point of the chord of the circle $x^2+y^2-6x+4y-12=0$ drawn parallel to the tangent at $(-1,1)$ and at a distance of one unit from the tangent is

The point $(5, -7)$ lies outside the circle:

The circumference of a circle passing through the point $(4,6)$ with two normals represented by $2x - 3y + 4 = 0$ and $x + y - 3 = 0$ is (in $\pi$)

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:

If $P(\frac{\pi}{3})$ and $Q(\frac{2\pi}{3})$ represent two points on the circle $x^2+y^2-4x+6y-12=0$ in parametric form,then the length of the chord $PQ$ is