Question

A tangent is drawn to each of the circles
$x^2+y^2=a^2,x^2+y^2=b^2$
Show that if the two tangents are mutually perpendicular, the locus of their point of intersection is a circle concentric with the given circles.

Answer 1

Any tangent to first circle is x cos $\alpha$ + y sin $\alpha$ = a as its distance from centre (0 , 0) is equal to radius a. Any tangent to x${}^2$ + y${}^2$ = b${}^2$ but perpendicular to above is obtained by replacing a by $\alpha$ - 90${}^{\circ }$ and its equation is
x cos ($\alpha$ - 90${}^{\circ }$) + y sin ($\alpha$ - 90${}^{\circ }$) = b
or x cos(90${}^{\circ }$ - $\alpha$) y sin(90${}^{\circ }$ - $\alpha$) = b
or x sin $\alpha$ - y cos $\alpha$ = b.
Locus of the point of intersection of these tangents as in part (a)is x${}^2$ + y${}^2$ = a${}^2$ + b${}^2$ which is a circle concentric with the given circles.