Local Behaviour of a Node of a Plane Curve In Miranda’s book “Algebraic Curves and Riemann Surfaces”, in page 70, it says

At a node, we locally have a curve of the form $xy=0$; the nearby curve looks locally like $xy=t$ for some small parameter $t$. Topologically, as $t$ approaches zero, a small circle (homeomorphic to $S^1$) is becoming contracted to the node point.

I am a bit confused with how a circle appears. 
 A: You have to look at the complex curves $C_t$ defined by $xy = t$ and $xy=0$.
For nonzero $t$, this is a genus $0$ curve, so its projective closure is homeomorphic to a sphere. For $t=0$, this will be homeomorphic to a sphere where we pinched one of its circles into a point.
In order to see properly this we have to compare the curves together with a family of suitable homeomorphisms. Namely, we would like if the trajectory of a point, as $t$ moves down towards $0$, converges to some point on one of the two axis.
Now let me define, for $t \in (0;1)$ the homeomorphism $\phi_t : C_1 \to C_t$ the following way :
If $|x| \ge 2$, then $\phi_t(x,y) = (x,ty)$.
If $|y| \ge 2$, then $\phi_t(x,y) = (tx,y)$.
If $\max(|x|,|y|) \le 1+t$, then $\phi_t(x,y) = (x\sqrt t, y\sqrt t)$.
If $|x|>1$ and $1+t \le |x|$, then $\phi_t(x,y) = (x\sqrt{|x|-1},yt/\sqrt{|x|-1})$.
If $|y|>1$ and $1+t \le |y|$, then $\phi_t(x,y) = (xt/\sqrt{|y|-1},y\sqrt{|y|-1})$.  
As you can see, a point $(x,y)$ is first pushed towards $(0,0)$ until it reaches a certain threshold upon which it is then slowly projected onto one of the two axis.
If we define $\phi_0(x,y) = \lim_{t \to 0} \phi_t(x,t)$, we get
If $|x| > 2$ then $\phi_0(x,y) = (x,0)$
If $|x| > 1$ then $\phi_0(x,y) = (x\sqrt{|x|-1},0)$
If $|y| > 1$ then $\phi_0(x,y) = (0,y\sqrt{|y|-1})$
If $|y| > 2$ then $\phi_0(x,y) = (0,y)$ 
and finally, if $|x|=|y|=1$ then $\phi_0(x,y) = (0,0)$.
$\phi_0$ is not an homeomorphism to $C_0$ because it is no longer injective, as all the points on $C_1$ with $|x|=|y|=1$ get pushed to $(0,0)$.
This subset is a circle (it is parametrised by the complex unit circle, with $u \mapsto (u,1/u)$), so looking at things through this particular family of maps, you see a circle collapses into a single point at $t=0$.  
If you work with the projective completions, this is a sphere where you pinch a circle down to a point. After resolving the singularity, that point splits in two, this disconnects the surface into two disjoint spheres, so two disjoint Riemann spheres (the $x$-axis and the $y$-axis)

