Determine the equation of the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ such that it has the least area but contains the circle $(x-1)^2+y^2=1$ Determine the equation of the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ such that it has the least area but contains the circle $(x-1)^2+y^2=1$

Since the area of ellipse is $A=\pi ab\Rightarrow A^2=\pi^2a^2b^2$ and the circle and the ellipse touch each other internally. This much I can visualise, but how to find out $a$ and $b$ from this?
 A: Firstly, define $f(x):=b\sqrt{1-\frac{x^2}{a^2}}$ and $g(x):=\sqrt{1-(x-1)^2}$ such that $f$ and $g$ represent the ellipse and the circle respectively in the upper half of the coordinate system.
In order to guarantee that the ellipse contains the circle, we need to have: 
$$
f(x)≥g(x)\iff b\sqrt{1-\frac{x^2}{a^2}}≥\sqrt{1-(x-1)^2} \iff b^2-\frac{b^2}{a^2}x^2≥2x-x^2\iff \\
\left(1-\frac{b^2}{a^2}\right)x^2-2x+b^2≥0
$$
In the last inequality, we have a quadratic function, which needs to be nonnegative everywhere, thus the discriminant $D$ has to be nonpositive:
$$
D=4-4\left(1-\frac{b^2}{a^2}\right)b^2≤0\iff 1≤b^2-\frac{b^4}{a^2} \iff \frac{b^4}{a^2}≤b^2-1
$$
We can see that $0<b^2-1\iff 1<b$, therefore we can divide by $b^2-1$ without any changes:
$$
a^2≥\frac{b^4}{b^2-1}\iff a≥\frac{b^2}{\sqrt{b^2-1}}
$$
This last inequality is sufficient and necessary for the given condition. 
Therefore, we have:
$$
A=\pi ab≥\frac{\pi b^3}{\sqrt{b^2-1}}
$$
Thus, the minimum of $A$ has to be greater than or equal to the minimum of $\frac{\pi b^3}{\sqrt{b^2-1}}$. This minimum can be found as follows:
Define $h(b):=\frac{\pi b^3}{\sqrt{b^2-1}}$.
$$
h'(b)=\frac{\pi b^2\left(2b^2-3\right)}{\left(b^2-1\right)^{\frac{3}{2}}} \implies \left(h'(b)=0\iff b=\sqrt{\frac{3}{2}}\right)
$$
$b=0$ is impossible because $b>1$. Therefore, the minimum has to be at $b=\sqrt{\frac{3}{2}}$ so we have $\min A≥h\left(\sqrt{\frac{3}{2}}\right)=\frac{\pi 3\sqrt{3}}{2}$. This minimum can indeed be achieved for $a=\frac{3}{\sqrt 2}$ and $b=\sqrt{\frac{3}{2}}$, for which we also have:
$$
a=\frac{3}{\sqrt 2}≥\frac{\sqrt{\frac{3}{2}}^2}{\sqrt{\sqrt{\frac{3}{2}}^2-1}}=\frac{b^2}{\sqrt{b^2-1}}
$$
Thus, the equation of the ellipse has to be:
$$
\frac{2x^2}{9}+\frac{2y^2}{3}=1
$$
A: Means ellipse and circle touches each other. So $\displaystyle \frac{x^2}{a^2}+\frac{y^2}{b^2} = 1$ and $(x-1)^2+y^2 = 1$
So Ellipse and Circle touches each other. So we will solve these two equations.
$\displaystyle \frac{x^2}{a^2}+\frac{1-(x-1)^2}{b^2} = 1\Rightarrow b^2x^2+a^2\left[1-(x-1)^2\right] = a^2b^2$
so $\displaystyle b^2x^2+a^2\left[1-(x^2+1-2x)\right]=a^2b^2$
So $\displaystyle b^2x^2+a^2-a^2x^2-a^2+2a^2x-a^2b^2 =0$
So $(b^2-a^2)x^2+2a^2x-a^2b^2$
So Discriminat of above equation is $=0$(Bcz Courve touches each other)
So $\displaystyle 4a^4+4(b^2-a^2)\cdot (a^2b^2) =0$
So $\displaystyle a^2+(b^2-a^2)b^2=0\Rightarrow a^2+b^4-a^2b^2=0$
