question_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 :   JEE Main  Solved  Paper-2014

A) \[\frac{\sqrt{3}}{\sqrt{2}}\]                                        

B) \[\frac{\sqrt{3}}{2}\]

C) \[\frac{1}{2}\]                                   

D) \[\frac{1}{4}\]

Correct Answer: D

Solution :

Equation of C is\[(x-1)2+(y-1)2=1\]?(1) Also let \[(0,y)\equiv (0,k)\] then equation of T is \[{{x}^{2}}+{{(y-k)}^{2}}={{k}^{2}}\]                                        ?(2) From the figure and equations (1) and (2) we get \[{{(1+k)}^{2}}={{1}^{2}}+{{(1-k)}^{2}}\]\[\Rightarrow \]\[k=\frac{1}{4}\] Hence radius of T is\[\frac{1}{4}.\]