<s>[INST] Find the equation of a circle tangent to a circle and x-axis, with center on a certain line. Find the equation of the circle, which is tangent to the circle $x^2 + y^2=4$ and $x-axis$, with center on the line $x=4$.
So far I have:$$(x-h)^2 + (y-k)^2=r^2$$
Since the center is on the line $x=4$, then the center is $(4;k)$.
$\therefore ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~(x-4)^2 + (y-k)^2=r^2$ 
And because the circle touches the $x-axis$, $r=k$.
$\therefore~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~(x-4)^2 + (y-k)^2=k^2$
Now, I am struggling to use $x^2 + y^2=4$ to find $k$.
Please help.
 [/INST] 
$$x^2+y^2=0$$
this circle have radius $=2$center$=(0,0)$ 
So, in $\triangle OO'A$ $$(OO')^2=(O'A)^2+(OA)^2\tag{Pythagoras theorem}$$
$$(OB+BO')^2=(O'A)^2+(OA)^2$$ $\therefore$ $$(2+k)^2=k^2+4^2$$ $$4+k^2+4k=k^2+16$$ $\therefore$ $$k=3$$
Hope it helps!!!  
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