To find the locus of a point 
A point moves such that the sum of its distances from the coordinate axes is equal to its distance from the circle $x^2+y^2=4$.

I tried and got this equation for the locus:
$$x+y=\sqrt{x^2+y^2}-2$$
Should I replace $x+y$ with $|x|+|y|$ because distances are absolute? In that case, how will I represent $y$ as an explicit function of $x$?
Thanks in advance.

Not only should you replace $x$ and $y$ with $|x|$ and $|y|$, you should also take the absolute value of the right-hand side because it represents a distance too:
$$|x|+|y|=\left\vert\sqrt{x^2+y^2}-2\right\vert$$
To solve this for $y$, we can first restrict ourselves to non-negative $x$ and $y$, since substituting $x\to-x$ or $y\to-y$ in a valid solution yields another valid solution. With this restriction we can remove the modulus signs around $x$ and $y$. The triangle inequality implies that $x+y>\sqrt{x^2+y^2}-2$, so when removing the modulus signs in the RHS we must have
$$x+y=2-\sqrt{x^2+y^2}$$
which leads to
$$2-(x+y)=\sqrt{x^2+y^2}$$
$$4-4(x+y)+x^2+2xy+y^2=x^2+y^2$$
$$4-4(x+y)+2xy=0$$
$$2-2(x+y)+xy=0$$
$$(x-2)y+2(1-x)=0$$
$$y=\frac{2(x-1)}{x-2},x\le1$$
Mirroring this across the $x$- and $y$-axes we get the final equation for $y$ in terms of $x$:
$$y=\pm\frac{2(|x|-1)}{|x|-2},|x|\le1$$
Here is a plot of the locus asked for by the question, with $x^2+y^2=4$ added as a reference.

