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.

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

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

Given equation of circle,

$$x^2+y^2=16$$

$\therefore$ Radius $=4$ and centre $=(0,0)$

Now, perpendicular from $(0,0)$ to line $y=\sqrt{3} x+k=$ Radius of the circle

$$\left|\frac{0-0+k}{\sqrt{3+1}}\right|=4$$

Since the distance from the point $(m, n)$ to the line $A x+B y+k=0$ is $d=\left|\frac{A m+B n+C}{A^2+B^2}\right|$

$$\begin{array}{lr} \Rightarrow & \pm \frac{k}{2}=4 \\ \therefore & k= \pm 8 \end{array}$$

Given equation of the circle is

$$\begin{aligned} & x^2+y^2-6 x+12 y+15=0 \\ & \therefore \quad 2 g=-6 \Rightarrow g=-3 \\ & 2 f=12 \Rightarrow f=6 \\ & \text { and } \quad c=15 \\ & \therefore \quad \text { Centre }=(-g,-f)=(3,-6) \end{aligned}$$

So, the centre of the required circle will be $(3,-6)\quad$. [since, the circles are concentric]

Radius of the given circle

$$\begin{aligned} & =\sqrt{g^2+f^2-c} \\ & =\sqrt{9+36-15}=\sqrt{30} \end{aligned}$$

Let radius of the required circle $=r_1$

$\therefore \quad 2 \times$ Area of the given circle $=$ Area of the required circle

$$\begin{aligned} &\begin{aligned} & \Rightarrow \quad 2\left[\pi(\sqrt{30})^2\right]=\pi r_1^2 \\ & \Rightarrow \quad 60=r_1^2 \\ & \Rightarrow \quad r_1=\sqrt{60} \\ & \therefore \quad \sqrt{g^2+f^2-c}=\sqrt{60} \\ & \Rightarrow \quad 9+36-c=60 \\ & \Rightarrow \quad c=-15 \end{aligned}\\ &\text { So, the required equation of circle is } x^2+y^2-6 x+12 y-15=0 \text {. } \end{aligned}$$