The centre$(s)$ of the circle$(s)$ passing through the points $(0, 0)$ and $(1, 0)$ and touching the circle $x^2 + y^2 = 9$ is/are:

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 two chords,each bisected by the $x$-axis,can be drawn to the circle $2(x^2 + y^2) - 2ax - by = 0$ $(a \ne 0, b \ne 0)$ from the point $P(a, b/2)$,then:

Let $P(3 \cos \alpha, 2 \sin \alpha)$, $\alpha \neq 0$, be a point on the ellipse $\frac{x^2}{9} + \frac{y^2}{4} = 1$, $Q$ be a point on the circle $x^2 + y^2 - 14x - 14y + 82 = 0$, and $R$ be a point on the line $x + y = 5$ such that the centroid of the triangle $PQR$ is $(2 + \cos \alpha, 3 + \frac{2}{3} \sin \alpha)$. Then the sum of the ordinates of all possible points $R$ is:

If the portion of the line $lx + my = 1$ falling inside the circle ${x^2} + {y^2} = {a^2}$ subtends an angle of $45^\circ$ at the origin,then

Given: $A$ circle $2x^2 + 2y^2 = 5$ and a parabola $y^2 = 4\sqrt{5}x$.
Statement-$1$: An equation of a common tangent to these curves is $y = x + \sqrt{5}$.
Statement-$2$: If the line $y = mx + \frac{\sqrt{5}}{m} (m \neq 0)$ is their common tangent,then $m$ satisfies $m^4 - 3m^2 + 2 = 0$.