The perpendicular distance from centre $(3,-4)$ to the line is, $d=\left|\frac{15-48-12}{\sqrt{25+144}}\right|=\frac{45}{13}$

So, the required equations of the circle is $(x-3)^2+(y+4)^2=\left(\frac{45}{13}\right)^2$.

Given equations of line are

$$\begin{aligned} y & =x+2 \quad \text{... (i)}\\ 3 y & =4 x \quad \text{... (ii)} \\ 2 y & =3 x \quad \text{... (iii)} \end{aligned}$$

From Eqs. (i) and (ii),

$$\begin{aligned} & \frac{4 x}{3}=x+2 \\ \Rightarrow \quad & 4 x=3 x+6 \end{aligned}$$

$$\begin{aligned} &\begin{aligned} & \Rightarrow x=6 \end{aligned}\\ &\text { On putting } x=6 \text { in Eq. (i), we get } \end{aligned}$$

$y=8$

$\therefore$ Point, $A=(6,8)$

From Eqs. (i) and (iii),

$$ \begin{aligned} & \frac{3 x}{2}=x+2 \\ \Rightarrow\quad & 3 x=2 x+4 \Rightarrow x=4 \end{aligned}$$

$$\begin{array}{rlrl} & \text { When } & x =4 \text {, then } y=6 \\ \therefore & \text { Point, } B =(4,6) \\ & \text { From Eqs. (ii) and (iii) } & x_1 =0, y=0 \\ &\text { Now, } & C =(0,0) \end{array}$$

Let the equation of circle is

$$x^2+y^2+2 g x n+2 f y+c=0$$

Since, the points $A(6,8), B(4,6)$ and $C(0,0)$ lie on this circle.

Since, the points $A(6,8), B(4,6)$ and $C(0,0)$ lie on this circle.

$$\begin{array}{rlrl} & 36+64+12 g+16 f+c =0 \\ \Rightarrow & 12 g+16 f+c =-100 \quad \text{... (iv)}\\ \text { and } & 16+36+8 g+12 f+c =0 \\ \Rightarrow & 8 g+12 f+c =-52 \quad \text{... (v)}\\ \Rightarrow & c =0\quad \text{... (vi)} \end{array}$$

From Eqs. (iv), (v) and (vi),

$$ \begin{aligned} & 12 g+16 f =-100 \\ \Rightarrow \quad & 3 g+4 f+25 =0 \\ \Rightarrow \quad & 2 g+3 f+13 =0 \\ \Rightarrow \quad & \frac{g}{+52-75} =\frac{f}{50-39}=\frac{1}{9-8} \\ \Rightarrow \quad & \frac{g}{-23} =\frac{f}{11}=\frac{1}{1} \\ \Rightarrow \quad & g =-23, f=11 \end{aligned}$$

$$\begin{aligned} &\text { So, the equation of circle is }\\ &\begin{aligned} & \quad x^2+y^2-46 x+22 y+0 =0 \\ &\Rightarrow \quad x^2+y^2-46 x+22 y=0 \end{aligned} \end{aligned}$$

Let equation of the ellipse be $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$.

$$\begin{array}{llrl} \therefore & 29 =6 \text { and } 2 b=4 \\ \Rightarrow & a =3 \text { and } b=2 \end{array}$$

We know that,

$$\begin{aligned} &\begin{aligned} c^2 & =a^2-b^2=(3)^2-(2)^2 \\ & =9-4=5 \Rightarrow c=\sqrt{5} \\ \therefore \quad \text { Length of string } & =A C^{\prime}+C^{\prime} C+A C \\ & =a+c+2 c+a c \\ & =2 a+2 c=6+2 \sqrt{5} \end{aligned}\\ &\therefore \quad \text { Distances between the pins }=2 \sqrt{5}=C C^{\prime} \end{aligned}$$

$$\begin{aligned} &\text { Given that, foci of the ellipse are }(0, \pm b e)\\ &\begin{array}{ll} \because & b e \equiv 1 \\ \therefore & \text { Length of minor axis, } 2 a=1 \Rightarrow a=1 / 2 \end{array} \end{aligned}$$

$$\begin{aligned} &\begin{aligned} e^2 & =1-\frac{a^2}{b^2} \\ \Rightarrow \quad(b e)^2 & =b^2-a^2 \Rightarrow 1=b^2-\frac{1}{4} \\ \Rightarrow \quad 1+\frac{1}{4} & =b^2 \Rightarrow \frac{5}{4}=b^2 \end{aligned}\\ &\text { So, the equation of ellipse is }\\ &\frac{x^2}{\frac{1}{4}}+\frac{y^2}{5 / 4}=1 \Rightarrow \frac{4 x^2}{1}+\frac{4 y^2}{5}=1 \end{aligned}$$

Given that, focus at $F(-1,-2)$ and directrix is $x-2 y+3=0$

Let any point on the parabola be $(x, y)$.

$$\begin{array}{ll} \therefore & \quad P F=\left|\frac{x-2 y+3}{\sqrt{1+4}}\right| \\ \Rightarrow & (x+1)^2+(y+2)^2=\frac{(x-2 y+3)^2}{5} \\ \Rightarrow & 5\left[x^2+2 x+1+y^2+4 y+4\right]=x^2+4 y^2+9-4 x y-12 y+6 x \\ \Rightarrow & 4 x^2+y^2+4 x+32 y+16=0 \end{array}$$