Question

A circle is given by $x^2+(y-1{)}^2=1.$ Then the locus of the centre of the circle touching the given circle and the $x$-axis, is

• A. $\left\{\right(x,y)|x^2=4y\}\cup \left\{\right(x,y)|y\le 0\}$
• B. $\left\{\right(x,y)|x^2+(y-1{)}^2=4\}\cup \left\{\right(x,y)|y\le 0\}$
• C. $\left\{\right(x,y)|x^2=4y\}$
• D. $\left\{\right(x,y)|x^2=4y\}\cup \left\{\right(0,y)|y\le 0\}$

Answer 1

The correct option is D $\left\{\right(x,y)|x^2=4y\}\cup \left\{\right(0,y)|y\le 0\}$

Let the locus of circle be $(h,k)$ touching ${(y-1)}^2+x^2=1$ and x-axis
Clearly from figure distance between O and A is always $1+\left|k\right|$
$\sqrt{{(h-0)}^2+{(k-1)}^2}=1+\left|k\right|\Rightarrow h^2+k^2-2k+1=1+k^2+2\left|k\right|\Rightarrow h^2=2\left|k\right|+2k\Rightarrow x^2=2\left|y\right|+2y$
where $\left|y\right|=\{\begin{array}{c}y,y\ge 0\\ -y,y<0\end{array}$
$\therefore x^2=2y+2y,y\ge 0$
and $x^2=-2y+2y,y<0\Rightarrow x^2=4y,y\ge 0$
and $x^2=0$ when $y<0$
$\therefore \{(x,y):x^2=4y,y\ge 0\}\cup \{(0,y):y<0\}$