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

Answer 1

Answer: (B)

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

Let $y=mx+r\sqrt{m^2+1}$ be a tangent to circle $x^2+y^2=a^2,C=(0,0),r=a$

$y=mx+a\sqrt{m^2+1}$

Taking only + sign

$⟹\frac{-mx}{a\sqrt{m^2+1}}+\frac{y}{a\sqrt{m^2+1}}=1$

$⟹\frac{x}{\frac{a\sqrt{m^2+1}}{-m}}+\frac{y}{a\sqrt{m^2+1}}=1$

Area of $\frac{x}{a}+\frac{y}{b}=1$ with axes $\frac{1}{2}|ab|$

$Area=\frac{1}{2}\times \frac{(a\sqrt{m^2+1}{)}^2}{|-m|}=a^2$

$\frac{a\sqrt{m^2+1}}{m}=\pm 2a^2$

$m^2+1\pm 2m=0$

$m=1,m=-1$

Tangent equation is

$y=\pm x\pm a\sqrt{1+1}$

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