All circles of the plane form a three-parametric family $$(x-a)^2+(y-b)^2 \;=\; r^2.$$ The parametres $a,\,b,\,r$ may be eliminated by using successive differentiations, when one gets $$x-a+(y-b)y' \;=\; 0,$$ $$1+y'^{\,2}+(y-b)y'' = 0,$$ $$3y'y''+(y-b)y''' \;=\; 0.$$ The two last equations allow to eliminate also $b$ , yielding the differential equation of all circles of the plane: $$(1+y'^{\,2})y'''-3y'y''^{\,2} \;=\; 0$$ It is of order three, corresponding the number of parametres.