Question

Find the locus of the centre of a circle which touches the x-axis and cuts $x^2+y^2-20x-10y+100=0$ orthogonally. What curve does it represent and determine its data.

Answer 1

Let the circle touching the x-axis be
$(x-h{)}^2+(y-k{)}^2=k^2$
or $x^2+y^2-2hx-2ky+k^2=0.....\left(1\right)$
It cuts $x^2+y^2-20x-10y+100=0$ orthogonally.
$\therefore 2h\left(10\right)+2k\left(5\right)=h^2+100$
$\therefore$ Locus of centre $(h,k)$ is
$x^2-20x+100=10y$
or $(x-10{)}^2=10y$
Above equation represents a parabola with vertex at $(10,0)$ and $L.R.=10$ and axis is along the line $x-10=0$ and tangent at vertex begin x-axis i.e. $y=0$.