$$\begin{aligned} \text{Given points are}\quad x & =\frac{2 a t}{1+t^2} \text { and } y=\frac{a\left(1-t^2\right)}{1+t^2} \\ \because\quad x^2+y^2 & =\frac{4 a^2 t^2}{\left(1+t^2\right)^2}+\frac{a^2\left(1-t^2\right)^2}{\left(1+t^2\right)^2} \end{aligned} $$ $$ \Rightarrow \quad \frac{1}{a^2}\left(x^2+y^2\right)=\frac{4 t^2+1+t^4-2 t^2}{\left(1+t^2\right)^2}$$

$$\begin{aligned} &\begin{array}{ll} \Rightarrow & \frac{1}{a^2}\left(x^2+y^2\right)=\frac{t^4+2 t^2+1}{\left(1+t^2\right)^2} \\ \Rightarrow & \frac{1}{a^2}\left(x^2+y^2\right)=\frac{\left(1+t^2\right)^2}{\left(1+t^2\right)^2} \end{array}\\ &\Rightarrow \quad x^2+y^2=a^2 \text {, which is a required circle. } \end{aligned}$$

Let the equation of circle is $$x^2+y^2+2 g x+2 f y=0\quad\text{... (i)}$$

Since, this circle passes through the points $A(0,0), B(a, 0)$ and $C(0, b)$.

$$\begin{aligned} \therefore \quad & a^2+2 a g=0 \quad\text{... (ii)}\\ \text{and}\quad & b^2+2 b f=0\quad \text{... (iii)} \end{aligned}$$

From Eq. (ii), $a+2 g=0 \Rightarrow g=-a / 2$

From Eq. (iii), $b+2 f=0 \Rightarrow t=-b / 2$

Hence, the coordinates of the circle are $\left(\frac{a}{2}, \frac{b}{2}\right)$.

Given that, centre of the circle is $(1,2)$.

$$\because$$ Radius = 2

So, the equation of circle is

$$\begin{array}{rrr} & (x-1)^2+(y-2)^2 & =2^2 \\ \Rightarrow & x^2-2 x+1+y^2-4 y+4 & =4 \\ \Rightarrow & x^2-2 x+y^2-4 y+1 & =0 \\ \Rightarrow & x^2+y^2-2 x-4 y+1 & =0 \end{array}$$

Given lines,

$$\begin{array}{r} 3 x-4 y+4=0 \quad \text{... (i)}\\ 6 x-8 y-7=0 \\ \text{or}\quad 3 x-4 y-7 / 2=0\quad \text{... (ii)} \end{array}$$

It is clear that lines (i) and (ii) parallel.

Now, distance between them i.e.,

$$d=\left|\frac{4+7 / 2}{\sqrt{9+16}}\right|=\left|\frac{\frac{8+7}{2}}{5}\right|=3 / 2$$

$\therefore$ Distance between these line $=$ Diameter of these circle

$\therefore$ Diameter of the circle $=3 / 2$

and radius of the circle $=3 / 4$

Let $a$ be the radius of the circle. Then, the coordinates of the circle are $(-a,-a)$. Now, perpendicular distance from $C$ to the line $A B=$ Radius of the circle

$$\begin{aligned} & d=\left|\frac{-3 a+4 a+8}{\sqrt{9+16}}\right|=\left|\frac{a+8}{5}\right| \\ \because \quad & a= \pm\left(\frac{a+8}{5}\right) \\ & \text { Taking positive sign, } \quad a=\frac{a+8}{5} \\ & \Rightarrow \quad 5 a=a+8 \\ & \Rightarrow \quad 4 a=8 \Rightarrow a=2 \\ & \text { Taking negative sign, } \quad a=\frac{-a-8}{5} \\ & \Rightarrow \quad 5 a=-a-8 \\ & \Rightarrow \quad 6 a=-8 \Rightarrow a=-4 / 3 \end{aligned}$$

$$\begin{aligned} &\begin{array}{ll} \text { But } & a \neq-4 / 3 \\ \because & a=2 \end{array}\\ &\text { So, the equation of circle is }\\ &\begin{array}{rlrl} & (x+2)^2+(y+2)^2 =2^2 \quad [\because a=2]\\ \Rightarrow & x^2+4 x+4+y^2+4 y+4 =4 \\ \Rightarrow & x^2+y^2+4 x+4 y+4 =0 \end{array} \end{aligned}$$