Question

The equation of the tangents to the circle $x^2+y^2=a^2$, which makes a triangle of area $a^2$ sq. units with coordinate axes, is/are

• A. $x+y=a\sqrt{2}$
• B. $x-y=a\sqrt{2}$
• C. $2x+3y=2a\sqrt{3}$
• D. $2x-3y=2a\sqrt{3}$

Answer 1

Answer: (A), (B)

$x-y=a\sqrt{2}$

Let the tangent be of form $\frac{x}{x_1}+\frac{y}{y_1}=1$
and area of triangle formed by coordinate axes is:
$\left|\frac{1}{2}{x}_1{y}_1\right|=a^2\cdots \left(1\right)$
Rearranging equation of tangent, we get
$x{y}_1+y{x}_1-x_1{y}_1=0$

Applying condition of tangency
$\left|\frac{-x_1{y}_1}{\sqrt{x_1^2+y_1^2}}\right|=a$
$\Rightarrow x_1^2+y_1^2=\frac{x_1^2{y}_1^2}{a^2}$

Using equation $\left(1\right)$, we get
$x_1^2+{\left(\frac{2a^2}{x_1}\right)}^2=4a^2\Rightarrow t+\frac{4a^4}{t}=4a^2(t=x_1^2)\Rightarrow t^2-4a^{2}t+4a^4=0\Rightarrow t=2a^2\Rightarrow x_1^2=2a^2\Rightarrow x_1=\pm a\sqrt{2}\Rightarrow y_1=\pm a\sqrt{2}$

Hence, the equation of the tangents are
$x+y=a\sqrt{2}x+y=-a\sqrt{2}x-y=a\sqrt{2}x-y=-a\sqrt{2}$