For a circle $C\left(\left(a,b\right),R\right)$ of center $(a,b)$ and radius $R$, I've used a function that transforms the circle $C$ to the curve $\mathscr{C}$ given by $$ \mathscr{C} \ : \ \left(x^2-y^2-a\right)^2+\left(2xy-b\right)^2=R^2 $$ which does not look sympathetic. However, my conjecture is that if $(0,0) \in C\left(\left(a,b\right),R\right)$ hence the $\mathscr{C}$ is formed of one component and if it does not, then it has two components as shown below.
It looks like particular type of Cassini's Oval but I dont know how to show it. And it seems like the more $(0,0)$ is near the boundary, the more the curve tends to be a lemnisate and then split into two components.
How can I show that if $(0,0) \notin C\left(\left(a,b\right),R\right)$ then $\mathscr{C}$ is not connected and that if $(0,0) \in C\left(\left(a,b\right),R\right)$ then $\mathscr{C}$ is connected ? How can I show that $\mathscr{C}$ is a Cassini's oval ?
Switching to polar coordinates is helpful here. Let $(d,\psi)$ be the polar coordinates of the center $(a,b)$, and $(\rho,\theta)$ those of point $(x,y)$. By convention I'll assume that $d,\rho\ge 0$ and $\psi,\theta\in\mathbb R$.
Cassini oval
If you work out your equation of $\mathscr C$ a bit, you obtain the polar equation $$ \mathscr C : \rho^4-2\big(\sqrt d\big)^2\rho^2\cos\left( 2\left(\theta-\frac\psi 2\right) \right) = \big(\sqrt R\big)^4 - \big(\sqrt d\big)^4 $$ According to wikipedia, that is the polar equation of a Cassini oval with foci at $\left(\sqrt d, \frac\psi 2\right)$ and $\left(\sqrt d, \pi+\frac\psi 2\right)$, and whose constant distance product is $\sqrt R$.
Connectedness
Still according to wikipedia, you can derive the properties you want from the fact that you have a Cassini oval. The shape is fully determined by $e=\sqrt{\frac Rd}$, and the wiki page explains the shape of a Cassini oval depending on whether $e<1$, $e=1$, or $e>1$, which corresponds to whether your original circle $C$ contains the origin or not.
Since I'm not familiar with Cassini ovals, I'm personally a bit unsatisfied with the lack of proof in wikipedia's description of the shape. If you want an alternative, you can still derive the property you want by examining the polar equation. For instance when $R\ge d$, it's not too hard to see that the domain enclosed by $\mathscr C$ is star-shaped with respect to the origin, thus implying that the boundary $\mathscr C$ must be connected.