Question

Find the equation of tangents to the circle $x^2+y^2=a^2$ which makes with axes a triangle of area $a^2$.

Answer 1

Any tangent to the circle $x^2+y^2=a^2$ is
$y=mx\pm a\sqrt{1+m^2}$.
Its intercepts on the axes are
$OA=h=\pm \left(a/m\right)\sqrt{1+m^2}$
and $k=\pm a\sqrt{(1+m^2)}$.
Area of the triangle formed by the tangent and the axes will be $\frac{1}{2}hk=a^2$ given.
$\therefore \pm \frac{1}{2}\cdot \frac{a}{m}\sqrt{1+m^2}\cdot a\sqrt{1+m^2}=a^2$
or $(1+m^2)=\pm 2m$ or $1+m^2\pm 2m=0$
or $(1\pm m^2)=0\therefore m=\pm 1$.
Hence the required tangents are $y=\pm x\pm a\sqrt{2}$.