Suppose two perpendicular tangents can be drawn from the origin to the circle $x^2+y^2-6 x-2 p y+17=0$, for some real $p$. Then, $|p|$ is equal to

$\text { Let } $S$ \text { be the circle in the } xy \text {-plane defined by the equation } x ^2+ y ^2=4 \text {. }$

($1$) 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

$(A)$ $x+y=4$ $(B)$ $(x-4)^2+(y-4)^2=16$ $(C)$ $(x-4)(y-4)=4$ $(D)$ $x y=4$

($2$) 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

$(A)$ $(x+y)^2=3 x y$ $(B)$ $x^{2 / 3}+y^{2 / 3}=2^{4 / 3}$ $(C)$ $x^2+y^2=2 x y$ $(D)$ $x^2+y^2=x^2 y^2$

Give the answer or quetion ($1$) and ($2$)

Length of the tangent from $({x_1},{y_1})$ to the circle ${x^2} + {y^2} + 2gx + 2fy + c = 0$ is

If the line $3x + 4y - 1 = 0$ touches the circle ${(x - 1)^2} + {(y - 2)^2} = {r^2}$, then the value of $r$ will be

The line $x\cos \alpha + y\sin \alpha = p$will be a tangent to the circle ${x^2} + {y^2} - 2ax\cos \alpha - 2ay\sin \alpha = 0$, if $p = $

At which point on $y$-axis the line $x = 0$ is a tangent to circle ${x^2} + {y^2} - 2x - 6y + 9 = 0$