Question

Find differential equation of all circles in the first quadrant which touch the co-ordinate axis.

Answer 1

Let the equation of the circle be $(x-h{)}^2+(y-h{)}^2=h^2$

$2(x-h)+2(y-h)y^{{}^{\prime }}=0$

$x+y{y}^{{}^{\prime }}=(1+y^{{}^{\prime }})h$

$\frac{x+y{y}^{{}^{\prime }}}{1+y^{{}^{\prime }}}=h$

Putting it back into the circle equation, we get

$(x-y{)}^2(1+y^{{}^{\prime }}{)}^2=(x+y{y}^{{}^{\prime }}{)}^2$