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

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

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

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

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

Correct Answer: D

Solution :

In \[\Delta ABC\] \[AB=y+1\], \[AC=1-y\], \[BC=1\]

\[A{{B}^{2}}=A{{C}^{2}}+B{{C}^{2}}\]

\[{{(y+1)}^{2}}={{(1-y)}^{2}}+{{(1)}^{2}}\]

\[{{(1+y)}^{2}}-{{(1-y)}^{2}}=1\]

\[4y=1\Rightarrow y=\frac{1}{4}\]Hence, radius of circle \[T=\frac{1}{4}\]