check that the unit disk of $\Bbb C$ is a subset of union of circles Let $\Bbb D$ be the unit disk of $\Bbb C$ and $a＞0$. Denote $S=\cup_{0≤r≤1}\{r^2+ar\sqrt{1-r^2}e^{i(\theta-\alpha)}:\theta,\alpha\in[0,2\pi]\}$.
Then $\Bbb D \subset S$ if and only if $$-1\in S_r:=\{x\in \Bbb C:|x–r^2|=ar\sqrt{1-r^2}\}$$ for some $r\in [0,1]$.
If $\Bbb D \subset S$, it is easy to check that there exists $r\in [0,1]$ such that $$-1\in S_r:=\{x\in \Bbb C:|x–r^2|=ar\sqrt{1-r^2}\}.$$
But if $-1\in S_r:=\{x\in \Bbb C:|x–r^2|=ar\sqrt{1-r^2}\}$
for some $r\in [0,1]$, we have $|-1-r_1^2|=ar_1\sqrt{1-r_1^2}\}$ for some $r_1\in [0,1]$.
How to prove that for any $\lambda$ with $|\lambda|≤1$, there exists $r\in[0,1]$ such that $$|\lambda–r^2|=ar\sqrt{1-r^2}$$.
 A: 
we have $|-1-r_1^2|=ar_1\sqrt{1-r_1^2}$ for some $r_1\in [0,1]$

Let $\,x = r_1^2 \in \mathbb R\,$, then squaring the equality:
$$
(-1-x)^2 = a^2x(1-x) \quad\iff\quad (a^2+1)x^2 - \left(a^2 -2\right) x + 1 = 0
$$
Since real roots are known to exist, the discriminant of the quadratic must be non-negative:
$$
0 \le \Delta = (a^2-2)^2-4(a^2+1)=a^2(a^2-8) \quad\iff\quad a^2 \ge 8
$$

prove that for any $\lambda$ with $|\lambda|≤1$, there exists $r\in[0,1]$ such that $|\lambda–r^2|=ar\sqrt{1-r^2}$

Let $\,x = r^2 \in \mathbb R\,$, then taking squared magnitudes on both sides:
$$
(\lambda-x)(\bar \lambda - x) = a^2x(1-x) \quad\iff\quad (a^2+1)x^2 - \left(a^2 + 2\text{Re}(\lambda)\right) x+ |\lambda|^2 = 0
$$
The latter is a real quadratic $\,p(x)\,$ with $\,p(0)=|\lambda|^2 \ge 0 \,$, $\,p(1)=|1-\lambda|^2 \ge 0\,$ and a minimum attained at $\,x_m = \frac{a^2 + 2\text{Re}(\lambda)}{2(a^2+1)} \in [0,1]\,$. The cases $\,\lambda \in \{0, 1\}\,$ are easily verified, otherwise the inequalities are strict $\,p(0), p(1) \gt 0\,$, and the condition for a real root in $\,[0,1]\,$ is $\,p(x_m) \le 0\,$:
$$
\begin{align}
4(a^2+1)\,p(x_m) &= \left(a^2+2\text{Re}(\lambda)\right)^2 - 2\left(a^2+2\text{Re}(\lambda)\right)^2 + 4(a^2+1)|\lambda|^2
\\ &= -\left(a^2+2\text{Re}(\lambda)\right)^2 + 4(a^2+1)|\lambda|^2
\\ &\le -(a^2-2)^2+4(a^2+1)
\\ &= -a^2(a^2-8) 
\\ &\le 0
\end{align}
$$
