Croatia_2018

Let the given circle be $x^2+y^2-4y+3=0$. Rewrite it in standard form:

$x^2+(y^2-4y)+3=0$ $x^2+(y-2{)}^2-4+3=0$ $x^2+(y-2{)}^2-1=0$ $x^2+(y-2{)}^2=1$

So, the circle has centre $(0,2)$ and radius $1$.

Let the centre of the required circle be $(h,k)$ and its radius be $r$.

Since the circle is tangent to the $x$-axis, its distance from the $x$-axis is $r$, so $k=r$.

Since the circle is externally tangent to the given circle, the distance between their centres is equal to the sum of their radii:

$\sqrt{(h-0{)}^2+(k-2{)}^2}=r+1$

But $k=r$, so:

$\sqrt{h^2+(r-2{)}^2}=r+1$

Square both sides:

$h^2+(r-2{)}^2=(r+1{)}^2$ $h^2+r^2-4r+4=r^2+2r+1$

Subtract $r^2$ from both sides:

$h^2-4r+4=2r+1$ $h^2-4r+4-2r-1=0$ $h^2-6r+3=0$

Recall $k=r$, so:

$h^2-6k+3=0$

Therefore, the locus of the centres $(h,k)$ is:

$h^2-6k+3=0$