Let equation of circle be

$$\begin{array}{ll} & (x-h)^2+(y-k)^2=r^2 \\ \Rightarrow \quad & (x-h)^2+(y-k)^2=9\quad \text{... (i)} \end{array}$$

Given that, centre $(h, k)$ lies on the line

$$y=x-1 i . e ., k=h-1\quad \text{... (ii)}$$

Now, the circle passes through the point $(7,3)$.

$$\begin{array}{lr} \therefore & (7-h)^2+(3-k)^3=9 \\ \Rightarrow & 49-14 h+h^2+9-6 k+k^2=9 \\ \Rightarrow & h^2+k^2-14 h-6 k+49=0\quad \text{... (iii)} \end{array}$$

On putting $k=h-1$ in Eq. (iii), we get

$$\begin{array}{rrr} & h^2+(h-1)^2-14 h-6(h-1)+49=0 \\ \Rightarrow & h^2+h^2-2 h+1-14 h-6 h+6+49=0 \\ \Rightarrow & 2 h^2-22 h+56=0 \\ \Rightarrow & h^2-11 h+28=0 \\ \Rightarrow & h^2-7 h-4 h+28=0 \\ \Rightarrow & h(h-7)-4(h-7)=0 \\ \Rightarrow & (h-7)(h-4)=0 \end{array}$$

$\therefore\qquad h=4,7$

When $h=7$, then $k=7-1=6$

$\therefore$ Centre $(7,6)$

When $h=4$, then $k=3$

$\therefore$ Centre $-(4,3)$

So, the equation of circle when centre $(7,6)$, is

$$\begin{array}{rrr} & (x-7)^2+(y-6)^2 & =9 \\ \Rightarrow & x^2-14 x+49+y^2-12 y+36 & =9 \\ \Rightarrow & x^2+y^2-14 x-12 y+76 & =0 \end{array}$$

When centre $(4,3)$, then the equation of the circle is

$$\begin{aligned} & (x-4)^2+(y-3)^2=9 \\ & \Rightarrow \quad x^2-8 x+16+y^2-6 y+9=9 \\ & \Rightarrow \quad x^2+y^2-8 x-6 y+16=0 \end{aligned}$$

(i) Given that, directrix $=0$ and focus $=(6,0)$

So, the equation of the parabola

$$\begin{array}{rlrl} & (x-6)^2+y^2 =x^2 \\ \Rightarrow & x^2+36-12 x+y^2 =x^2 \\ \Rightarrow & y^2-12 x+36 =0 \end{array}$$

(ii) Given that, vertex $=(0,4)$ and focus $=(0,2)$

So, the equation of parabola is

$$\begin{array}{rlrl} & \sqrt{(x-0)^2+(y-2)^2} =|y-6| \\ \Rightarrow & x^2+y^2-4 y+4 =y^2-12 y+36 \\ \Rightarrow & x^2-4 y+12 y-32 =0 \\ \Rightarrow & x^2+8 y-32 =0 \\ \Rightarrow & x^2 =32-8 y \end{array}$$

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

So, the equation of parabola is $\sqrt{(x+1)^2+(y+2)^2}=\left|\frac{x-2 y+3}{\sqrt{1+4}}\right|$

$$\begin{array}{lrl} \Rightarrow & x^2+2 x+1+y^2+4 y+4 & =\frac{1}{5}\left[x^2+4 y^2+9-4 x y-12 y+6 x\right] \\ \Rightarrow & 4 x^2+4 x y+y^2+4 x+32 y+16 & =0 \end{array}$$

Let the coordinates of the point be $(x, y)$, then according to the question,

$$\begin{array}{ll} & \sqrt{(x-3)^2+y^2}+\sqrt{(x-9)^2+y^2}=12 \\ \Rightarrow & \sqrt{(x-3)^2+y^2}=12-\sqrt{(x-9)^2+y^2} \end{array}$$

$$\begin{aligned} &\text { On squaring both sides, we get }\\ &\begin{array}{llrl} & x^2-6 x+9+y^2 =144+\left(x^2-18 x+81+y^2\right)-24 \sqrt{(x-9)^2+y^2} \\ \Rightarrow & 12 x-216 =-24 \sqrt{(x-9)^2+y^2} \\ \Rightarrow & x-18 =-2 \sqrt{(x-9)^2+y^2} \\ \Rightarrow & x^2-36 x+324 =4\left(x^2-18 x+81+y^2\right) \\ \Rightarrow & 3 x^2+4 y^2-36 x =0 \end{array} \end{aligned}$$

Let the point be $P(x, y)$.

$\therefore$ Distance from $(0,4)=\sqrt{x^2+(y-4)^2}$

So, the distance from the line $y=9$ is $\left|\frac{y-9}{\sqrt{1}}\right|$

$$\begin{array}{lr} \therefore & \sqrt{x^2+(y-4)^2}=\frac{2}{3}\left|\frac{y-9}{1}\right| \\ \Rightarrow & x^2+y^2-8 y+10=\frac{4}{9}\left(y^2-18 y+81\right) \\ \Rightarrow & 9 x^2+9 y^2-72 y+144=4 y^2-72 y+324 \\ \Rightarrow & 9 x^2+5 y^2=180 \end{array}$$

Let the points be $P(x, y)$.

$\therefore$ Distance of $P$ from $(4,0) \sqrt{(x-4)^2+y^2}\quad \text{... (i)}$

and the distance of $P$ from $(-4,0) \sqrt{(x+4)^2+y^2}\quad \text{... (ii)}$

Now,

$$\begin{aligned} & \sqrt{(x+4)^2+y^2}-\sqrt{(x-4)^2+y^2}=2 \\ & \sqrt{(x+4)^2+y^2}=2+\sqrt{(x-4)^2+y^2} \end{aligned}$$

On squaring both sides, we get

$$\begin{array}{rlrl} & x^2+8 x+16+y^2 =4+x^2-8 x+16+y^2+4 \sqrt{(x-4)^2+y^2} \\ \Rightarrow & 16 x-4 =4 \sqrt{(x-4)^2+y^2} \\ \Rightarrow & 4(4 x-1) =4 \sqrt{(x-4)^2+y^2} \\ \Rightarrow & 16 x^2-8 x+1=x^2+16-8 x+y^2 \\ \Rightarrow & 15 x^2-y^2 =15 \text { which is a parabola. } \end{array}$$