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A circle \mathrm{\mathrm{C}_2} passes through the centre of the circle \mathrm{\mathrm{C}_1}; \mathrm{x^2+y^2=2} and gets bisected at its point of intersection with \mathrm{\mathrm{C}_1}. The radius of \mathrm{\mathrm{C}_2} is equal to 

 

Option: 1

1 units


Option: 2

2 units


Option: 3

\sqrt{2}


Option: 4

none of these


Answers (1)

best_answer

Let the radius of the circle be r.

Clearly AP=BP=OP=r

\mathrm{\begin{aligned} & \Rightarrow A P^2+O P^2=O A^2=2 \\ & \Rightarrow A P=1 \Rightarrow r=1 \end{aligned}}

 

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