Question

Form the differential equation of the family of circles in the first quadrant, which touches the coordinate axes.

Answer 1

If a circle touches coordinate axis then the center will be at $(a,a)$ and will have a radius of $a$

Equation of circle is ${(x-a)}^2+{(y-a)}^2=a^2$

Differentiating this for one time, we get

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

From here, we find $a$ which is $\frac{2x+2y{y}^{{}^{\prime }}}{2(1+y^{{}^{\prime }})}$

This can be plugged in main equation and we can get required differential equation

Equation of circle is ${(x-\frac{2x+2y{y}^{{}^{\prime }}}{2(1+y^{{}^{\prime }})})}^2+{(y-\frac{2x+2y{y}^{{}^{\prime }}}{2(1+y^{{}^{\prime }})})}^2={\left(\frac{2x+2y{y}^{{}^{\prime }}}{2(1+y^{{}^{\prime }})}\right)}^2$