Question

The locus of the point of intersection of two perpendicular tangents to the circle $x^2+y^2=a^2$ is

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

Answer 1

The correct option is C $x^2+y^2=2a^2$

Equation of a tangent to the circle $x^2+y^2=a^2$ is y = mx + a$\sqrt{1+m^2}$ $\to$ (1)
Equation of a tangent to the circle $x^2+y^2=a^2$ and perpendicular to (1) is
y = $(-\frac{1}{m})$ x + a $\sqrt{1+(-\frac{1}{m})}$ $\Rightarrow$ y = - $\frac{x}{m}$ + $\frac{a\sqrt{1+m^2}}{m}$ $\Rightarrow$ my = - x + a$\sqrt{1+m^2}\to \left(2\right)$
From (1), y - mx = a$\sqrt{1+m^2}$
From (2), my + x = a$\sqrt{1+m^2}$
Squaring and adding the above questions, we get $(y-mx{)}^2+(my+x{)}^2=a^2(1+m^2)+a^2(1+m^2)$
$\Rightarrow (1+m^2)x^2+(1+m^2)y^2=2a^2(1+m^2)$
$\Rightarrow x^2+y^2=2a^2$ which is the required locus.