Question

If a circle be drawn so as always to touch a given straight line and also a given circle' prove that the locus of its centre is a parabola.

Answer 1

Taking the center of the fixed circle as origin the axes being perpendicular and parallel to the given line respectively, the equation of the given circle will be
$x^2+y^2=r^2.....1$
and of the line $x=a....2$
Let center be $(h,k)$. So its equation may be written as
$x^2+y^2-2hx-2ky+c=\mathrm{0....3}$
Radius of $3$ is $\sqrt{h^2+k^2-c}$
If $1$ touches $3$ , the distance between the centers must be equal to the sum of radii
So,
$\sqrt{h^2+k^2}=\sqrt{(h^2+k^2-c)}+r......4$
If $3$ touches $2$ length of perpendicular from $(h,k)$ on $2$ must be equal to the radius
$h-a=\sqrt{h^2+k^2-c}......5$
$\sqrt{h^2+k^2}=h-a+r$
$h^2+k^2=h^2+a^2+r^2-2ha+2rh-2ar$
Locus is;-
$y^2=2r(r-a)+{(a-r)}^2$
which is a parabola