Inscribed and circumscribed circles

Question:
Two circles are given
$$\begin{eqnarray} & & x^2+y^2-6ax+2ay+20a-10 = 0  \\ & & x^2+y^2 = 4 \end{eqnarray}$$
Find $a$ so these circles are inscribed or circumscribed.

Answer:
Equation (1) can be arranged as follows:
$$\begin{eqnarray} & & x^2 -6ax + (3a)^2 -(3a)^2 + y^2 +2ay + (a)^2 -(a)^2 = 10 - 20a  \nonumber \\ & & \rightarrow x^2 -6ax + (3a)^2 + y^2 +2ay + (a)^2  = 10 - 20a + (3a)^2 + (a)^2 \nonumber  \\ & & \rightarrow (x - 3a)^2 + (y + a)^2 = 10a^2 -20a +10  \nonumber \\ & & \rightarrow (x - 3a)^2 + (y + a)^2 = 10(a-1)^2 \end{eqnarray}$$
Therefore, (2) has the center $(0, 0)$ and the radius $2$. Circle (3) has the center $(3a, -a)$ and radius $\sqrt{10}|a-1|$. The distance between those centers is
$$d = \sqrt{(3a-0)^2+(-a-0)^2}=\sqrt{10a^2}=\sqrt{10}|a|$$
If $a=1$, the radius becomes zero. Thus, $a \neq 1$. Let $r_1$ and $r_2$ be radii of two circles. In general, when $d=r_1 + r_2$, they are circumscribed. When $d=|r_1 - r_2|$, they are inscribed. Then, these conditions can be combined as $d = |r_1 \pm r_2|$. Now consider the case in $a>1$. Use the definition above:
$$\begin{equation} \sqrt{10}|a| = | \sqrt{10}(a-1) \pm 2 | \end{equation}$$
Square both sides and solve for $a$.
$$\begin{equation} a = \frac{5 \pm \sqrt{10}}{10} < 1 \end{equation}$$
The result is contradicted. Consider when $a<1$.
$$\begin{equation} \sqrt{10}|a| = | \sqrt{10}(1-a) \pm 2 | \end{equation}$$
Likewise, solve for $a$.
$$\begin{equation} a = \frac{5 \pm \sqrt{10}}{10} \end{equation}$$
which is the proper result.

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