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

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

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

Given equation of the circle is

$$\begin{aligned} & x^2+y^2-4 x-6 y+11=0 \\ \therefore & 2 g=-4 \text { and } 2 f=-6 \end{aligned}$$

So, the centre of the circle is $(-g,-f) i . e$., $(2,3)$.

Since, the mid-point of $A B$ is $(2,3)$.

Then,

$$\begin{aligned} 2 & =\frac{3+x_1}{2} \\ \Rightarrow\quad 4 & =3+x_1 \\ \therefore\quad x_1 & =1 \\ \text{and}\quad 3 & =\frac{4+y_1}{2} \\ \Rightarrow\quad 6 & =4+y_1 \Rightarrow y_1=2 \end{aligned}$$

So, the coordinates of other end of the diameter will be $(1,2)$.

Given that, centre of the circle is $(1,-2)$ and the circle passing through the lines

$$\begin{array}{r} 3 x+y=14 \quad \text{.... (i)}\\ \text{and}\quad2 x+5 y=18\quad \text{... (ii)} \end{array}$$

From Eq. (i) $y=14-3 x$ put in Eq. (ii), we get

$$\begin{gathered} 2 x+70-15 x=18 \\ -13 x=-52 \Rightarrow x=4 \end{gathered}$$

Now, $x=4$ put in Eq. (i), we get

$$12+y=14 \Rightarrow y=2$$

Since, point $(4,2)$ lie on these lines also lies on the circle.

$$\begin{aligned} \therefore \quad \text { Radius of the circle } & =\sqrt{(4-1)^2+(2+2)^2} \\ & =\sqrt{9+16}=5 \end{aligned}$$

$$\begin{aligned} &\text { Now, equation of the circle is }\\ &\begin{array}{lrr} & (x-1)^2+(y+2)^2=5^2 \\ \Rightarrow & x^2-2 x+1+y^2+4 y+4=25 \\ \Rightarrow & x^2+y^2-2 x+4 y-20=0 \end{array} \end{aligned}$$