Every triple of real numbers $(x,y,r)$, where $r>0$, represents a circle with center $(x,y)$ and radius $r$.
Suppose we have a infinite sequence of such triples $(x_i,y_i,r_i)$, that converges to $(x_L,y_L,r_L)$, with $r_L>0$.
For every $i$, define $d_i$ as the Hausdorff distance between the circle represented by $(x_i,y_i,r_i)$ and the circle represented by the limit $(x_L,y_L,r_L)$.
Is this true that the sequence $d_i$ converges to 0? Intuitively it seems true, but what is the way to prove it?
I assume by "circle" you mean just the circumference, not the disc.
Consider two such circles $C_1(x_1,y_1,r_1)$ and $C_2(x_2,y_2,r_2)$. The points at angle $\theta$ on the two circles are $$ p_1(\theta) = \left[\begin{matrix}x_1\\y_1\end{matrix}\right] + r_1\left[\begin{matrix}\cos\theta\\\sin\theta\end{matrix}\right] \qquad\text{and}\qquad p_2(\theta) = \left[\begin{matrix}x_2\\y_2\end{matrix}\right] + r_2\left[\begin{matrix}\cos\theta\\\sin\theta\end{matrix}\right] $$ and the distance between those points satisfies $$ \|p_1(\theta)-p_2(\theta)\| \le \left\|\left[\begin{matrix}x_1-x_2\\y_1-y_2\end{matrix}\right]\right\| + |r_1-r_2| \le |x_1-x_2|+|y_1-y_2|+|r_1-r_2| $$ The existence of such a point $p_2(\theta)$ on the one circle for every point $p_1(\theta)$ on the other, and vice versa, shows that $$ d(C_1,C_2) \le |x_1-x_2|+|y_1-y_2|+|r_1-r_2| $$ This implies that the map sending $(x,y,r)\in\mathbb{R}^3$ to the corresponding circle is continuous (even $1$-Lipschitz wrt the $\ell_1$ norm on $\mathbb{R}^3$), and so convergence of the tuples implies convergence of the circles, as desired.