$$\begin{array}{r} & \text { Given lines are } & 2 x-3 y-5 =0 \quad \text{... (i)}\\ & \text { and } & 3 x-4 y-7 =0 \\ & \text { From Eqs. (i) and (ii), } & \frac{x}{21-20} =\frac{y}{-15+14}=\frac{1}{-8+9} \\ \Rightarrow & \frac{x}{1} =\frac{y}{-1}=\frac{1}{+1} \\ \Rightarrow & x = \pm 1, y=-1 \end{array}$$

Since the intersection point of these lines will be coordinates of the circle i.e., coordinates of the circle as $(1,-1)$.

Let the radius of the circle is $r$.

$$\begin{aligned} \text{Then,}\quad \pi r^2 & =154 \\ \Rightarrow \quad \frac{22}{7} \times r^2 & =154 \\ \Rightarrow\quad r^2 & =\frac{154 \times 7}{22} \\ \Rightarrow\quad r^2 & =\frac{14 \times 7}{2} \Rightarrow r^2=49 \end{aligned}$$

So, the equation of circle is

$$\begin{aligned} & (x-1)^2+(y+1)^2 =49 \\ \Rightarrow \quad & x^2-2 x+1+y^2+2 y+1 =49 \\ \Rightarrow \quad & x^2+y^2-2 x+2 y =47 \end{aligned}$$

Let the general equation of the circle is

$$x^2+y^2+2 g x+2 f y+c=0\quad \text{... (i)}$$

Since, this circle passes through the points $(2,3)$ and $(4,5)$.

$$\begin{array}{lrl} \therefore & 4+9+4 g+6 f+c & =0 \\ \Rightarrow & 4 g+6 f+c & =-13 \quad\text{... (ii)}\\ \text { and } & 16+25+8 g+10 f+c & =0 \\ \Rightarrow & 8 g+10 f+c & =-41\quad \text{... (iii)} \end{array}$$

Since, the centre of the circle $(-g,-f)$ lies on the straight line $y-4 x+3=0$

i.e., $\quad +4 g-f+3=0\quad \text{... (iv)}$

From Eq. (iv), $4 g=f-3$

On putting $4 \mathrm{~g}=f-3$ in Eq. (ii), we get

$$\begin{aligned} & f-3+6 f+c=-13 \\ \Rightarrow & 7 f+c=10\quad \text{.... (v)} \end{aligned}$$

From Eqs. (ii) and (iii),

$$\begin{gathered} 8 g+12 f+2 c=-26 \\ 8 g+10 f+c=-41 \\ \frac{-\quad-\quad-\qquad\quad+}{2 f+c=15}\quad \text{... (vi)} \end{gathered}$$

From Eqs. (ii) and (vi),

$$ \begin{gathered} 7 f+c=-10 \\ 2 f+c=15 \\ \frac{-\quad-\quad-}{5 f=-25} \end{gathered}$$

$$\begin{aligned} \therefore\quad f & =-5 \\ \text{Now,}\quad c & =10+15=25 \\ \text{From Eq. (iv),}\quad 4 g+5+3 & =0 \\ \Rightarrow\quad g & =-2 \end{aligned}$$

From Eq. (i), equation of the circle is $x^2+y^2-4 x-10 y+25=0$.

Given centre of the circle is $(3,-1)$.

$$\begin{aligned} \text{Now,}\quad& O P=\left|\frac{6+5+18}{\sqrt{4+25}}\right|=\frac{29}{\sqrt{29}}=\sqrt{29} \\ \text{In } \triangle O P B,\quad & O B^2=O P^2+P B^2 \quad {[\because A B=6 \Rightarrow P B=3]} \\ \Rightarrow \quad & O B^2=29+9 \Rightarrow O B^2=38 \end{aligned}$$

So, the radius of circle is $\sqrt{38}$,

$\therefore$ Equation of the circle with radius $r=\sqrt{38}$ and centre $(3,-1)$ is

$$\begin{array}{r} \Rightarrow \quad & (x-3)^2+(y+1)^2=38 \\ \Rightarrow \quad & x^2-6 x+9+y^2+2 y+1=38 \\ &x^2+y^2-6 x+2 y=28 \end{array}$$

Let the coordinates of centre of the required circle are $(h, k)$, then the centre of another circle is $(1,2)$.

Radius $=\sqrt{1+4+20}=5$

So, it is clear that $P$ is the mid-point of $C_1 C_2$.

$$\begin{array}{r} \therefore \quad & 5=\frac{1+h}{2} \Rightarrow h=9 \\ \text { and } \quad & 5=\frac{2+k}{2} \Rightarrow k=8 \end{array}$$

So, the equation of and required circle is

$$\begin{array}{r} & (x-9)^2+(y-8)^2 =25 \\ \Rightarrow \quad & x^2-18 x+81+y^2-16 y+64 =25 \\ \Rightarrow \quad & x^2+y^2-18 x-16 y+120 =0 \end{array}$$

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

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