What is the degree of the differential equation of all circles touching both the coordinate axes in the first quadrant?

General equation of circle, $[x-h]^2+(y-k)^2=r^2$ As the circle touches both the coordinate axes in first quadrant, $\begin{aligned} & \therefore \text { Centre, } c(h, k)=(r, r) \\ & (x-r)^2+(y-r)^2=r^2 \qquad {...\mathrm{(i)}}\\ & x^2-2 x r+y^2-2 y r+r^2=0 \\ & 2 x-2 r+2 y y_1-2 r y_1=0 \\ & r\left(1+y_1\right)=x+y y_1 \\ & \therefore r=\frac{x+y y_1}{1+y_1} \end{aligned}$ Substituting the value of $r$ in eqn.(1) and solving that eqn. we finally get. $\left(x+y y_1\right)^2=(x-y)^2\left(1+y_1^2\right)$ Max power of $y_1=$ Degree $=2$