A circle of radius 1 unit touches positive x-axis and positive y-axis at A and B, respectively. A variable line passing through origin intersects the circle in two points D and E. If the area of the triangle DEB is maximum when the slope of the line is m, then find the value of m^–2

$\begin{array}{l}From the given data equation of circle be {(x-1)}^2+{(y-1)}^2=1\\ \Rightarrow x^2+y^2-2x-2y+1=0\to \left(1\right)\\ Equation of variable line be y=mx\to \left(2\right)\\ By solving \left(1\right) and \left(2\right) \\ (1+m^2)x^2-2x(1+m)+1=0\\ \text{∴ Length }\mathrm{DE}=\sqrt{\frac{8\mathrm{m}}{1+{\mathrm{m}}^2}}\\ Area of ∆DEB\left(A\right)=\frac{1}{2}\times DE\times dis\tan ce of B from DE\\ \Rightarrow A^2 =\frac{1}{4}\left(\frac{8\mathrm{m}}{1+{\mathrm{m}}^2}\right)\left(\frac{1}{1+m^2}\right)\\ A^2 =\frac{2m}{{(1+m^2)}^2}\\ \Rightarrow A=\frac{\sqrt{2m}}{1+m^2}\\ \frac{dA}{dm}=\frac{1-3m^2}{\sqrt{3}{(1+m^2)}^2}=0\\ \Rightarrow m=\pm \sqrt{3}\\ \Rightarrow \frac{{\mathrm{d}}^2\mathrm{A}}{{\mathrm{dm}}^2}<0,\text{ if }\mathrm{m}=\frac{1}{\sqrt{3}}\\ \therefore Area is maximum when \mathrm{m}=\frac{1}{\sqrt{3}}\\ Now \frac{1}{m^2}=3\end{array}\exists$