The differential equation of co-axal system of circles ${\mathrm{x}}^2+{\mathrm{y}}^2-4+\mathrm{\lambda } (2\mathrm{x}+\mathrm{y}-5)=0$ is
 

$\begin{array}{rcl} & & {\mathrm{x}}^2+{\mathrm{y}}^2+\mathrm{\lambda }(2\mathrm{x}+\mathrm{y}-5)-0\\ & \Rightarrow & -\mathrm{\lambda }=\frac{\left({\mathrm{x}}^2+{\mathrm{y}}^2-4\right)}{2\mathrm{x}+\mathrm{y}-5}\end{array}$

on differentiating we get

$$\begin{gathered} 0=\frac{(2\mathrm{x}+\mathrm{y}-5) (2\mathrm{x}+2\mathrm{y}{\displaystyle \frac{\mathrm{dy}}{\mathrm{dx}}})-2+{\displaystyle \frac{\mathrm{dy}}{\mathrm{dx}}}\left({\mathrm{x}}^2+{\mathrm{y}}^2-4\right)}{{\left(2\mathrm{x}+\mathrm{y}-5\right)}^2} \\ \Rightarrow \left(2+\frac{\mathrm{dy}}{\mathrm{dx}}\right) \left({\mathrm{x}}^2+{\mathrm{y}}^2-4\right)-\left(2\mathrm{x}+\mathrm{y}-5\right) \left(2\mathrm{x}+2\mathrm{y}\frac{\mathrm{dy}}{\mathrm{dx}}\right)=0 \end{gathered}$$