Question

The locus of the point of intersection of the perpendicular tangents to the circles $x^2$ + $y^2$ = $a^2$, $x^2$ + $y^2$ = $b^2$ is

• A. + = +
• B. + = -
• C. + =
• D. + =

Answer 1

The correct option is A + = +

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