14. Given the family of circles equation $x^{2}+y^{2}-4 a x+2 a y+20 a-20=0$.
(1) Prove that for any real number $a$, all the circles pass through a fixed point;
(2) Find the value of $a$ for the circles that are tangent to $x^{2}+y^{2}=4$.
14. (1) The original equation can be transformed into $\left(x^{2}+y^{2}-20\right)+2 a(2 x-y-10)=0$. Let $\left\{\begin{array}{l}x^{2}+y^{2}-20=0, \\ 2 x-y-10=0,\end{array}\right.$ solving this gives $\left\{\begin{array}{l}x=4, \\ y=-2,\end{array}\right.$ thus both pass through the fixed point $(4,-2)$.
(2) Subtract the equations of the two circles, $-2 a x+a y+10 a-8=0$, which is the line equation that the intersection points of the two circles should satisfy. Clearly, the distance from $(0,0)$ to it is 2, i.e., $\frac{|10 a-8|}{\sqrt{5 a^{2}}}=2$ (it is easy to see that $a \neq 0$), thus $a=1 \pm \frac{\sqrt{5}}{5}$.