How many solutions ?

The system of equations can be rewritten as:

$\begin{cases} \begin{aligned} (x-2)^2 + y^2 & = (r+1)^2 & ...(1) \\ (x+1)^2 + y^2 & = (r+2)^2 & ...(2) \\ x^2 + y^2 & = (3-r)^2 & ...(3) \end{aligned} \end{cases}$

$\begin{aligned} (2)-(1): \quad 6x - 3 & = 2r + 3 \\ \implies r & = 3x - 3 & ...(4) \end{aligned}$

$\begin{aligned} (2)-(3): \quad 2x +1 & = 10r-5 \\ \implies 5r & = x + 3 & ...(5) \end{aligned}$

From $5\times(5)-(4): \ 14 r = 12 \implies r = \dfrac 67$ . We note that the system of equations are circles as follows:

$\begin{cases} \begin{aligned} (x-2)^2 + y^2 & = \left(\frac {13}7 \right)^2 & \small \blue{\text{Center at (2,0) and radius of } \frac {13}7} \\ (x+1)^2 + y^2 & = \left(\frac {20}7 \right)^2 & \small \blue{\text{Center at (-1,0) and radius of } \frac {20}7} \\ x^2 + y^2 & = \left(\frac {15}7 \right)^2 & \small \blue{\text{Center at (0,0) and radius of } \frac {15}7} \end{aligned} \end{cases}$

Plotting the three circles, we note that there are only $\boxed 2$ solutions at $(r,x,y) = \left(\dfrac 67, \dfrac 97, \pm \dfrac {12}7\right)$ .

Subtracting the first equation from the second, we get $r=3x-3$ . Subtracting the third equation from the first, we get $2r=3-x$ . From these two, we get $x=\dfrac{9}{7}, r=\dfrac{6}{7}$ . Substituting in the third equation we get $y=\pm {\dfrac{12}{7}}$ . Hence there are $\boxed 2$ equations in all, viz. $x=\dfrac{9}{7}, y=\dfrac{12}{7}, r=\dfrac{6}{7}$ and $x=\dfrac{9}{7}, y=-\dfrac{12}{7}, r=\dfrac{6}{7}$ .