Let the equation of circle be
$$
x^2+y^2+2 g x+2 f y+c=0
$$
where, centre $(-g,-f)$
The centre of given circle $x^2+y^2-20 x+4=0$ is $(10,0)$,
Condition of two circles cut.
$$
\therefore \begin{aligned}
2\left(g_1 g_2+f_1 f_2\right) & =c_1+c_2 \\
2(-g \times 10+0 \times(-f) & =c+4 \\
2(-10 g) & =c+4
\end{aligned}
$$
Also, circle touch the line $x=2$.
$\therefore$ The perpendicular distance from centre to the circle is equal to radius of the circle.
$$
\begin{aligned}
& \therefore \quad \frac{|-g-2|}{\sqrt{1}}=\sqrt{g^2+f^2-c} \\
& \Rightarrow \quad(g+2)=\sqrt{g^2+f^2-c} \\
& \Rightarrow \quad g^2+4+4 g=g^2+f^2-c \\
& \Rightarrow \quad f^2-4 g-c-4=0 \\
& \Rightarrow f^2-4 g+4+20 g-4=0 \\
& \Rightarrow \quad f^2+16 g=0
\end{aligned}
$$
Hence, the locus of $(-g,-f)$ is
$$
y^2-16 x=0 \Rightarrow y^2=16 x
$$