The locus of the centre of the circle, which cuts the circle $x^2+y^2-20 x+4=0$ orthogonally and touches the line $x=2$, is

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 $$ y^2-16 x=0 \Rightarrow y^2=16 x $$