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

Option: 1

\mathrm{ \left\{x^2=4 y, y \geq 0\right\} \cup\{(0, y), y<0\} }


Option: 2

\mathrm{ x^2=y}


Option: 3

\mathrm{ \mathrm{y}=4 \mathrm{x}^2 }


Option: 4

\mathrm{y}^2=4 \mathrm{x} \cup(0, \mathrm{y}) \mathrm{y} \in \mathrm{R}


Answers (1)

best_answer

Let the locus of centre of circle be \mathrm{(h, k)}  touching \mathrm{(y-1)^2+x^2=1}  and x-axis shown as clearly from figure distance between \mathrm{O} and \mathrm{A} is always \mathrm{1+|k|}  ie \mathrm{\sqrt{(h-0)^2+(k-1)^2}=1+|k| }
 

\begin{aligned} & \Rightarrow h^2+k^2-2 k+1=1+k^2+2|k| \\ & h^2=2|k|+2 k \Rightarrow x^2=2|y|+2 y \\ & x^2=2 y+2 y \quad y \geq 0 \\ & x^2=-2 y+2 y \quad y<0 \Rightarrow x^2=4 y(y \geq 0) \\ & x^2=0(y<0) \\ & \therefore\left\{(x, y) ; x^2=4 y(y \geq 0)\right\} \cup\{(0, y), y<0\} \end{aligned}

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Rakesh

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