# q.1 find the locus of the feet of the perpendicular drawn from the point (b,0) on tangents to the circle x2+y2 =a2.

the equation of the tangent to the circle
let P(h,k) be the locus of the feet of the perpendicular drawn from the point A(b,0).
therefore PA is perpendicular to the tangents
slope of PA = - 1/m
$\frac{k-0}{h-b}=-\frac{1}{m}\phantom{\rule{0ex}{0ex}}⇒m=\frac{b-h}{k}.......\left(3\right)$

and point (h,k) must lie on the tangent (2), therefore

now for locus replace h by x and k by y:
${\left[{y}^{2}-bx+{x}^{2}\right]}^{2}={a}^{2}.\left[{y}^{2}+\left(b-x{\right)}^{2}\right]$

hope this helps you

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