Let a and b be respectively the degree and order of the differential equation of the family of circles touching the lines$y^2 — x^2 = 0$ and lying in the first and second quadrant then

 Given, $a$ and $b$ be the respectively degree and family of circle touching the line $y{}^2-x^2=0$ and lying in first and second quadrants.
By symmetry and also by congruency
$\{y-x=0$ and $y+x=0\}$, of $∆OAC$ and $∆0BC$,
linter $\left(C\right)$ of the circles will lie on $y$-axis.
let $C=(0,\alpha )\Rightarrow OC=\alpha$
$\therefore$ the lines are perpendicular to each other. So,
$$\begin{gathered} \angle OCA={45}^{\circ } \\ \Rightarrow \mu =\alpha \cos 45°=\frac{\alpha }{\sqrt{2}} \end{gathered}$$
$\therefore$ Equation of family of $1$ circles $=\frac{\alpha }{\sqrt{2}}$.
Differentiating both sides, we have. $={\left(\frac{\alpha }{\sqrt{2}}\right)}^2$$....... \left(1\right)$
$\begin{array}{r}2x+2(y-\alpha )y^{\text{'}}=0 \left[y^{\text{'}}=\frac{dy}{dx}\right]\text{. }\end{array}$
$\alpha =y+\frac{{\displaystyle x}}{y}..........\left(2\right)$
from $\left(1\right),$ we obtain

$$\begin{gathered} x^2+\left(y-{\left(y+\frac{x}{y^2}\right)}^2={\left(y+\frac{x}{y^{\text{'}}}\right)}^2\right. \\ =2\left(x^2+\frac{x^2}{y^{\text{'}2}}\right)=y^2+\frac{x^2}{y^{\text{'}2}}+\frac{2xy}{y^{\text{'}}} \\ 2x^2+\frac{x^2}{y^{\text{'}2}}-y^2-\frac{2xy}{y^{\text{'}}}=0 \\ \Rightarrow 2x^2{y}^{\text{'}2}+x^2-y^2{y}^{\text{'}2}-2xy{y}^{\text{'}}=0\Rightarrow {\left(\frac{dy}{dx}\right)}^2\left(2x^2-y^2\right)-2xy\left(\frac{dy}{dx}\right)+k^2=0. \\ \text{ So, Degree}=2\text{ and order }=1 \mathrm{i},\mathrm{e}.,\mathrm{a}=2 \mathrm{and} \mathrm{b}=1 \end{gathered}$$