The line $2x - y + 1 = 0$ is tangent to the circle at the point $(2, 5)$ and the centre of the circle lies on $x - 2y = 4$. The radius of the circle is

Let $S$ be the circle in the $xy$-plane defined by the equation $x^2+y^2=4$.
$(1)$ Let $E_1, E_2$ and $F_1, F_2$ be the chords of $S$ passing through the point $P_0(1,1)$ and parallel to the $x$-axis and the $y$-axis,respectively. Let $G_1, G_2$ be the chord of $S$ passing through $P_0$ and having slope $-1$. Let the tangents to $S$ at $E_1$ and $E_2$ meet at $E_3$,the tangents to $S$ at $F_1$ and $F_2$ meet at $F_3$,and the tangents to $S$ at $G_1$ and $G_2$ meet at $G_3$. Then,the points $E_3, F_3$,and $G_3$ lie on the curve
$(A)$ $x+y=4$ $(B)$ $(x-4)^2+(y-4)^2=16$ $(C)$ $(x-4)(y-4)=4$ $(D)$ $xy=4$
$(2)$ Let $P$ be a point on the circle $S$ with both coordinates being positive. Let the tangent to $S$ at $P$ intersect the coordinate axes at the points $M$ and $N$. Then,the mid-point of the line segment $MN$ must lie on the curve
$(A)$ $(x+y)^2=3xy$ $(B)$ $x^{2/3}+y^{2/3}=2^{4/3}$ $(C)$ $x^2+y^2=2xy$ $(D)$ $x^2+y^2=x^2y^2$

If the length of the tangent from any point on the circle $(x-3)^2+(y+2)^2=5r^2$ to the circle $(x-3)^2+(y+2)^2=r^2$ is $16$ units, then the area between the two circles in sq. units is (in $\pi$)

The equation of a line inclined at an angle $\frac{\pi}{4}$ to the $X$-axis,such that the two circles $x^2 + y^2 = 4$ and $x^2 + y^2 - 10x - 14y + 65 = 0$ intercept equal lengths on it,is

The given circle $x^2 + y^2 + 2px = 0$,$p \in R$ touches the parabola $y^2 = 4x$ externally,then

If $y = m_{1}x + c_{1}$ and $y = m_{2}x + c_{2}$ with $m_{1} \neq m_{2}$ are two common tangents of the circle $x^{2} + y^{2} = 2$ and the parabola $y^{2} = x$,then the value of $8|m_{1}m_{2}|$ is equal to