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. $\sqrt{(h-0)^{2}+(k-1)^{2}}=1+|k|$,
$\Rightarrow h^{2}+k^{2}-2 k+1$
$=1+k^{2}+2|k|$
$\Rightarrow h^{2}=2|k|+2 k$
$\Rightarrow x^{2}=2|y|+2 y$
where $|y|=\left\{\begin{array}{cc}y, & y \geq 0 \\ -y, & y<0\end{array}\right.$
$\therefore x^{2}=2 y+2 y, y \geq 0$
and $ x^{2}=-2 y+2 y, y<0$
$\Rightarrow x^{2}=4 y$, when $y \geq 0$
and $ x^{2}=0$, when $y<0$
$\therefore \left\{(x, y): x^{2}=4 y\right.$, when $\left.y \geq 0\right\} \cup\{(0, y): y<0\}$