Question

The locus of the centre of circle which touches $(y-1{)}^2+x^2=1$ externally and also touches x-axis, is

• A. $\{x^2=4y,y\ge 0\}\cup \left\{\right(0,y),y<0\}$
• B. $x^2=y$
• C. $y=4x^2$
• D. $y^2=4x\cup (0,y),yϵR$

Answer 1

The correct option is A $\{x^2=4y,y\ge 0\}\cup \left\{\right(0,y),y<0\}$

Let the locus of centre of circle be (h, k) touching $(y-1{)}^2+x^2=1$ and X- axis shown as


Clearly, from figure,
Distance between C and A is always $1+|k|$,
i.e.$\sqrt{(h-0{)}^2+(k-1{)}^2}=1+|k|$
$\Rightarrow h^2+k^2-2k+1=1+k^2+2|k|\Rightarrow h^2=2|k|+2k\Rightarrow x^2=2|y|+2y$
where $|y|=\{\begin{array}{ccc}y& ,y\ge 0& \\ -y& ,& y<0\end{array}$
$\therefore x^2=2y+2y,y\ge 0$
and $x^2=-2y+2y,y<0$
$\Rightarrow x^2=4y,wheny\ge 0$
and $x^2=0$, when $y<0$
$\therefore \left\{\right(x,y):x^2=4y,wheny\ge 0\}\cup \left\{\right(0,y);y<0\}$