Therefore, given locus is a circle with center $$(1, 2)$$ and radius $$\sqrt 2 .$$
3
JEE Main 2014 (Offline)
MCQ (Single Correct Answer)
Let $$C$$ be the circle with centre at $$(1, 1)$$ and radius $$=$$ $$1$$. If $$T$$ is the circle centred at $$(0, y)$$, passing through origin and touching the circle $$C$$ externally, then the radius of $$T$$ is equal to
A
$${1 \over 2}$$
B
$${1 \over 4}$$
C
$${{\sqrt 3 } \over {\sqrt 2 }}$$
D
$${{\sqrt 3 } \over 2}$$
Explanation
Equation of circle $$C \equiv {\left( {x - 1} \right)^2} + {\left( {y - 1} \right)^2} = 1$$
Radius of $$T = \left| y \right|$$
$$T$$ touches $$C$$ externally
therefore,
Distance between the centers $$=$$ sum of their radii