Question

The equation of the tangent to the circle $x^2+y^2=a^2$, which makes a triangle of area $a^2$ with the coordinate axes, is

• A. $x\pm y=a\sqrt{2}$
• B. $x+y=\pm a\sqrt{2}$
• C. $x\pm y=2a$
• D. $x+y=\pm 2a$

Answer 1

The correct option is B $x+y=\pm a\sqrt{2}$

let the equation of the tangent be $\frac{x}{l}+\frac{y}{m}=1$
given that $\frac{lm}{2}=a^2$ ...(1)
now, radius = perpendicular distance

$a=\left|\frac{1}{\sqrt{\frac{1}{l^2}+\frac{1}{m^2}}}\right|$

$\Rightarrow \frac{l^2{+m}^2}{l^2{m}^2}=\frac{1}{a^2}$
$\Rightarrow l^2{+m}^2=4a^2$ ... from (1)
$\Rightarrow l^2+\frac{4a^4}{l^2}=4a^2$ ....from (1)
$\Rightarrow l=\pm \sqrt{2}a\Rightarrow m=\pm \sqrt{2}a$
Therefore, equation of tangent is $x+y=\pm a\sqrt{2}$