Find the condition for the chord lx+my = 1 of the circle ${\mathrm{x}}^2+{\mathrm{y}}^2={\mathrm{a}}^2$ to subtend a right angle at the origin.

The given circle is ${\mathrm{x}}^2+{\mathrm{y}}^2={\mathrm{a}}^2 ...\left(1\right)$

 and the given line is lx + my = 1 ... (2)
Let A and B be the point of intersection of given line and circle.
Homogenizing eq. (1) with eq (2), we get combined equation of $\overline{\mathrm{OA}}\text{ and }\overline{\mathrm{OB}}$
$\begin{array}{l}{\mathrm{x}}^2+{\mathrm{y}}^2-{\mathrm{a}}^2(1{)}^2=0\\ \Rightarrow {\mathrm{x}}^2+{\mathrm{y}}^2-{\mathrm{a}}^2(\mathrm{lx}+\mathrm{my}{)}^2=0\\ \Rightarrow {\mathrm{x}}^2+{\mathrm{y}}^2-{\mathrm{a}}^2\left({\mathrm{l}}^2{\mathrm{x}}^2+{\mathrm{m}}^2{\mathrm{y}}^2+2\mathrm{lmxy}\right)=0\\ \Rightarrow {\mathrm{x}}^2+{\mathrm{y}}^2-{\mathrm{a}}^2{\mathrm{l}}^2{\mathrm{x}}^2-{\mathrm{a}}^2{\mathrm{m}}^2{\mathrm{y}}^2-2\mathrm{almxy}=0\\ \Rightarrow \left(1-{\mathrm{a}}^2{\mathrm{l}}^2\right){\mathrm{x}}^2-2\mathrm{almxy}+\left(1-{\mathrm{a}}^2{\mathrm{m}}^2\right){\mathrm{y}}^2=0\dots \end{array}$
Equation (3) Represents pair of mutually perpendicular lines
$$\begin{gathered} \mathrm{then} \mathrm{a}+\mathrm{b}=0 \\ \Rightarrow 1-{\mathrm{a}}^2{\mathrm{l}}^2+1-{\mathrm{a}}^2{\mathrm{m}}^2=0 \\ \Rightarrow 2={\mathrm{a}}^2\left({\mathrm{l}}^2+{\mathrm{m}}^2\right) \end{gathered}$$
Is the required condition