Question

If a circle be drawn so as always to touch a given straight line and also a given circle externally then prove that the locus of its centre is a parabola. (given line and given circle are non intersecting)

Answer 1

Thus the locus is:-

$x^2+y^2=r^2+\left(\frac{(mh-k+c{)}^2}{m^2+1}\right)+2r\left|\frac{mh-k+c}{\sqrt{m^2}+1}\right|$

Let the circle be:

$x^2+y^2=r^2$

and the line be:-

$y=mx+c$

Here distance of line from origin is greater than $r$

Let the centre of circle be $(h,k)$. This circle is variable which touches the fixed circle. Let $r$ be radius of variable circle.

So, distance of point $(h,k)$ from origin is:-

$\sqrt{(h-0{)}^2+(k-0{)}^2}=r+r---\left(1\right)$

As $r$ is equal to distance of $(h,k)$ from $(0,0)$, so,

$r=\left|\frac{mh-k+c}{\sqrt{m^2+1}}\right|$

substituting in $\left(1\right)$

$\sqrt{n^2+k^2}=r+\left|\frac{mh-k+c}{\sqrt{m^2+1}}\right|$

$=n^2+k^2=r^2+\left(\frac{(mh-k+c{)}^2}{m^2+1}\right)+2r\left|\frac{mh-k+c}{\sqrt{m^2+1}}\right|$