Circle Ques 47
- 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
(2014 Main)
(a) $\frac{\sqrt{3}}{\sqrt{2}}$
(b) $\frac{\sqrt{3}}{2}$
(c) $\frac{1}{2}$
(d) $\frac{1}{4}$
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Answer:
Correct Answer: 47.(d)
Solution:
Formula:
- PLAN Use the property, when two circles touch each other externally, then distance between the centre is equal to sum of their radii, to get required radius.
Let the coordinate of the centre of $T$ be $(0, k)$.
Distance between their centre
$ \begin{aligned} k+1 & =\sqrt{1+(k-1)^{2}} \quad\left[\because C _1 C _2=k+1\right] \\ \Rightarrow \quad k+1 & =\sqrt{1+k^{2}+1-2 k} \end{aligned} $
$ \begin{aligned} & \Rightarrow & k+1 & =\sqrt{k^{2}+2-2 k} \\ & \Rightarrow & k^{2}+1+2 k & =k^{2}+2-2 k \\ & \Rightarrow & k & =\frac{1}{4} \end{aligned} $
So, the radius of circle $T$ is $k$, i. e. $\frac{1}{4}$.
BETA
