All circles of the plane form a threeparametric family

$${(xa)}^{2}+{(yb)}^{2}={r}^{2}.$$ 

The parametres $a,b,r$ may be eliminated by using successive differentiations^{}, when one gets

$$xa+(yb){y}^{\prime}=\mathrm{\hspace{0.33em}0},$$ 


$$1+{y}^{\prime \mathrm{\hspace{0.17em}2}}+(yb){y}^{\prime \prime}=0,$$ 


$$3{y}^{\prime}{y}^{\prime \prime}+(yb){y}^{\prime \prime \prime}=\mathrm{\hspace{0.33em}0}.$$ 

The two last equations allow to eliminate also $b$, yielding
the differential equation^{} of all circles of the plane:

$$(1+{y}^{\prime \mathrm{\hspace{0.17em}2}}){y}^{\prime \prime \prime}3{y}^{\prime}{y}^{\prime \prime \mathrm{\hspace{0.17em}2}}=\mathrm{\hspace{0.33em}0}$$ 

It is of three, corresponding the number of parametres.