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

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

Consider the equation of the ellipse is $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$.

$\therefore$ Length of major axis $=2 a$

Length of minor axis $=2 b$

and length of latusectum $=\frac{2 b^2}{a}$

Given that,

$$\begin{aligned} \frac{2 b^2}{a} & =\frac{2 b}{2} \\ \Rightarrow a & =2 b \Rightarrow b=a / 2 \\ \end{aligned}$$

We know that, $b^2 =a^2\left(1-e^2\right)$

$$\begin{array}{ll} \Rightarrow & \left(\frac{a}{2}\right)^2=a^2\left(1-e^2\right) \\ \Rightarrow & \frac{a^2}{4}=a^2\left(1-e^2\right) \\ \Rightarrow & 1-e^2=\frac{1}{4} \\ \Rightarrow & e^2=1-\frac{1}{4} \\ \therefore & e=\sqrt{\frac{3}{4}}=\sqrt{\frac{3}{2}} \end{array}$$

$$\begin{array}{rlrl} & \text { Given equation of ellipse, } & 9 x^2+25 y^2 =225 \\ \Rightarrow & \frac{x^2}{25}+\frac{y^2}{9} =1 \\ \Rightarrow & a=5, b =3 \\ & \text { We know that, } & b^2 =a^2\left(1-e^2\right) \end{array}$$

$$\begin{array}{l} \Rightarrow & 9 =25\left(1-e^2\right) \\ \Rightarrow & \frac{9}{25} =1-e^2 \\ \Rightarrow & e^2 =1-9 / 25 \\ \therefore & e=\sqrt{1-9 / 25} =\sqrt{\frac{25-9}{25}} \\ & =\sqrt{\frac{16}{25}}=4 / 5 \end{array}$$

$$\text { Foci }=( \pm a e, 0)=( \pm 5 \times 4 / 5,0)=( \pm 4,0)$$

Given that, eccentricity $=\frac{5}{8}$, i.e., $e=\frac{5}{8}$

Let equation of the ellipse be $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$,

Since the foci of this ellipse is ( $\pm \mathrm{ae}, 0$ ).

$\therefore \quad$ Distance between foci $=\sqrt{(a e+a e)^2}$

$$\begin{array}{lll} \Rightarrow & 2 \sqrt{a^2 e^2}=10 & {[\because \text { distance between its foci }=10]} \\ \Rightarrow & \sqrt{a^2 e^2}=5 \\ \Rightarrow & a^2 e^2=25 \\ \Rightarrow & a^2=\frac{25 \times 64}{25} \\ \therefore & a=8 \end{array}$$

We know that,

$$\begin{array}{ll} \Rightarrow & b^2=a^2\left(1-e^2\right) \\ \Rightarrow & b^2=64\left(1-\frac{25}{64}\right) \\ \Rightarrow & b^2=64\left(\frac{64-25}{64}\right) \\ & b^2=39 \end{array}$$

$\therefore \quad$ Length of latusrectum of ellipse $=\frac{2 b^2}{a}=2\left(\frac{39}{8}\right)=\frac{39}{4}$

Given that, $e=2 / 3$ and latusrectum $=5$ i.e.,

$$\begin{aligned} \frac{2 b^2}{a} & =5 \Rightarrow b^2=\frac{5 a}{2} \\ \text{We know that,}\quad b^2 & =a^2\left(1-e^2\right) \end{aligned}$$

$$\begin{array}{ll} \Rightarrow & \frac{5 a}{2}=a^2\left(1-\frac{4}{9}\right) \\ \Rightarrow & \frac{5}{2}=\frac{5 a}{9} \Rightarrow a=9 / 2 \Rightarrow a^2=\frac{81}{4} \\ \Rightarrow & b^2=\frac{5 \times 9}{2 \times 2}=\frac{45}{4} \end{array}$$

So, the required equation of the ellipse is $\frac{4 x^2}{81}+\frac{4 y^2}{45}=1$.