Get Answers to all your Questions

header-bg qa

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

Option: 1

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


Option: 2

\mathrm{\left\{(x, y): x^2+(y-1)^2=4\right\} \cup\{(x, y): y \leq 0\}}


Option: 3

\mathrm{\left\{(x, y): x^2=y\right\} \cup\{(0, y): y \leq 0\}}


Option: 4

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


Answers (1)

best_answer

Let the centre of circle C be (h, k).

Then as this circle touches axis of x its radius = |k| Also it touches the given
circle \mathrm{x^2+(y-1)^2=1}, centre (0, 1) and radius = 1, externally
Therefore, the distance between centres = sum of radii

\mathrm{\begin{aligned} & \Rightarrow \sqrt{(h-0)^2+(k-1)^2}=1+|k| \\ & \Rightarrow h^2+k^2-2 k+1=(1+|k|)^2 \\ & \Rightarrow h^2+k^2-2 k+1=1+2|k|+k^2 \Rightarrow h^2=2 k+2|k| \\ & \therefore \text { Locus of }(h, k) \text { is, } x^2=2 y+2|y| \\ & \text { Now if } y>0 \text {, it becomes } x^2=4 y \\ & \text { and if } y \leq 0 \text {, it becomes } x=0 \end{aligned}}

Combining the two, the required locus is

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

Posted by

seema garhwal

View full answer

JEE Main high-scoring chapters and topics

Study 40% syllabus and score up to 100% marks in JEE