question_answer

A  circle is given by \[{{x}^{2}}+{{(y-1)}^{2}}=\text{ }1,\] another circle C touches it externally and also the x-axis, then the locus of its centre is

A) \[\{(x,y):{{x}^{2}}=4y\}\cup \{(x,y):y\le 0\}\]

B) \[\{(x,y):{{x}^{2}}+{{(y-1)}^{2}}=4\}\cup \{(x,y):y\le 0\}\]

C) \[\{(x,y):{{x}^{2}}=y\}\cup \{(0,y):y\le 0\}\]

D) \[\{(x,y):{{x}^{2}}=4y\}\cup \{(0,y):y\le 0\}\]

Correct Answer: D

Solution :

Let the centre of circle C be (h, k). Then as this circle touches axis of\[x\], its radius \[=\left| k \right|\]

Also it touches the given circle \[{{x}^{2}}+{{(y-1)}^{2}}=1,\]

Centre (0, 1) radius 1, externally

Therefore, the distance between centres = sum of radii

\[\Rightarrow \]\[\sqrt{{{(h-0)}^{2}}+{{(k-1)}^{2}}}=1+\left| k \right|\]

\[\Rightarrow \]\[{{h}^{2}}+{{k}^{2}}-2k+1={{(1+\left| k \right|)}^{2}}\]

\[\Rightarrow \]\[{{h}^{2}}+{{k}^{2}}-2k+1=1+2\left| k \right|+{{k}^{2}}\]

\[\Rightarrow \]\[{{h}^{2}}=2k+2\left| k \right|\]

\[\therefore \]locus of \[(h,k)\] is, \[{{x}^{2}}=2y+2\left| y \right|\]

Now if \[y>0,\]it becomes\[{{x}^{2}}=4y\]

and if \[y\le 0,\]it becomes \[x=0\]

\[\therefore \]Combining the two, the required locus is

\[\{(x,y):{{x}^{2}}=4y\}\cup \{(0,y):y\le 0\}\]