Question

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

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

Answer 1

Answer: (A)

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 $|y|=\{\begin{array}{cc}y,& y\ge 0\\ -y,& y<0\end{array}$
$\therefore x^2=2y+2y,y\ge 0$
and $x^2=-2y+2y,y<0$
$\Rightarrow x^2=4y$, when $y\ge 0$
and $x^2=0$, when $y<0$
$\therefore \{(x,y):x^2=4y$, when $y\ge 0\}\cup \{(0,y):y<0\}$