What equation intersects only once with $f(x)=\sqrt{1-(x-2)^2}$ Being $f(x)=\sqrt{1-(x-2)^2}$ I have to know what linear equation only touches the circle once(only one intersection), and passes by $P(0,0)$.
So the linear equation must be $y=mx$ because $n=0$.
I have a system of 2 equations:
\begin{align}
y&=\sqrt{1-(x-2)^2}\\
y&=mx
\end{align}
So I equal both equations and I get \begin{align}
mx&=\sqrt{1-(x-2)^2}\\
m&=\frac{\sqrt{1-(x-2)^2}}{x}
\end{align}
$m$ can be put in the $y=mx$ equation, which equals to:
\begin{align}
y&=\left(\frac{\sqrt{1-(x-2)^2}}{x}\right)x\\
&=\sqrt{1-(x-2)^2}
\end{align}
But that equation has $\infty$ intersections, and I want only the equation who has $1$ interception.

What is the good way to know this? And how can it be calculated?


You were doing fine up through $mx=\sqrt{1-(x-2)^2}$. Starting there, square both sides to get $m^2x^2=1-(x-2)^2$, and rewrite this as $(m^2+1)x^2-4x+3=0$. Think of this as a quadratic in $x$; solving yields
$$x=\frac{4\pm\sqrt{16-12(m^2+1)}}{2(m^2+1)}=\frac{2\pm\sqrt{4-3(m^2+1)}}{m^2+1}\;,$$
which has two solutions or none unless $4-3(m^2+1)=0$, i.e., $m^2=\frac13$, and $m=\pm\frac1{\sqrt3}$. 
However, the equation $f(x)=\sqrt{1-(x-2)^2}$ describes only the upper half of the circle with centre $\langle 2,0\rangle$ and radius $1$, so the line through the origin must have positive slope in order to intersect the semicircle. Its equation is therefore $$y=\frac{x}{\sqrt3}\;.$$
