Find the pair of tangents from the origin the circle ${\mathrm{x}}^2+{\mathrm{y}}^2+2\mathrm{gx}+2\mathrm{fy}+\mathrm{c}=0$ and hence deduce a condition for these tangents to be perpendicular.

Given circle is
$\mathrm{S}={\mathrm{x}}^2+{\mathrm{y}}^2+2\mathrm{gx}+2\mathrm{fy}+\mathrm{c}=0$ origin is 0 (0,0)
Equation of pair of tangents from origin to
$\mathrm{S}=0\text{ is }{\mathrm{ss}}_{11}={\mathrm{s}}_1^2$

$\begin{array}{l}\Rightarrow \left({\mathrm{x}}^2+{\mathrm{y}}^2+2\mathrm{gx}+2\mathrm{fy}+\mathrm{c}\right)(0+\mathrm{c})\\ =[0+0+\mathrm{g}(\mathrm{x}+0+\mathrm{f}(\mathrm{y}+0)+\mathrm{c}){]}^2\\ \Rightarrow \mathrm{c}\left({\mathrm{x}}^2+{\mathrm{y}}^2\right)+2\mathrm{c}(\mathrm{gx}+\mathrm{fy})+{\mathrm{c}}^2=(\mathrm{gx}+\mathrm{fy}+\mathrm{c}{)}^2\\ \Rightarrow \mathrm{c}\left({\mathrm{x}}^2+{\mathrm{y}}^2\right)+2\mathrm{c}(\mathrm{gx}+\mathrm{fy})+{\mathrm{c}}^2\\ =(\mathrm{gx}+\mathrm{fy}{)}^2+{\mathrm{c}}^2+2\mathrm{c}(\mathrm{gx}+\mathrm{fy})\\ \Rightarrow {\mathrm{cx}}^2+{\mathrm{cy}}^2={\mathrm{g}}^2{\mathrm{x}}^2+{\mathrm{f}}^2{\mathrm{y}}^2+2\mathrm{gfxy}\\ \Rightarrow \left(\mathrm{c}-{\mathrm{g}}^2\right){\mathrm{x}}^2-2\mathrm{gfxy}+\left(\mathrm{c}-{\mathrm{f}}^2\right){\mathrm{y}}^2=0\end{array}$
It represents pair of tangents OA & OB compairs with ${\mathrm{ax}}^2+2\mathrm{hxy}+{\mathrm{by}}^2=0$
$\mathrm{a}=\mathrm{c}-{\mathrm{g}}^2,\mathrm{b}=\mathrm{c}-{\mathrm{f}}^2$
Tangents are mutually ${\perp }^{\text{er }}(\text{ ie }\mathrm{OA}\perp \mathrm{OB})$
If $\text{ cocff .of }{\mathrm{x}}^2+\text{ cocff of }{\mathrm{y}}^2=0$
$\begin{array}{l}\Rightarrow \mathrm{a}+\mathrm{b}=0\Rightarrow \mathrm{c}-{\mathrm{g}}^2+\mathrm{c}-{\mathrm{f}}^2=0\\ \Rightarrow {\mathrm{g}}^2+{\mathrm{f}}^2=2\mathrm{c}\end{array}$
Which is the condition for the tangents to be ${\perp }^{\mathrm{ar}}$