question_answer

The locus of the mid points of the chords of the  circle \[{{x}^{2}}+{{y}^{2}}=4\]which subtend a right angle at  the origin is                                                

A)  \[{{x}^{2}}+{{y}^{2}}=1\]                            

B)  \[{{x}^{2}}+{{y}^{2}}=2\]        

C)  \[x+y=1\]          

D)  \[x+y=2\]      

Correct Answer: B

Solution :

Let mid point of the chord AB is \[C({{x}_{1}},{{y}_{1}})\] In \[\Delta COB,\sin \frac{\pi }{4}=\frac{BC}{OB}\] \[\Rightarrow \]               \[\frac{1}{\sqrt{2}}=\frac{BC}{2}\] \[\Rightarrow \]               \[BC=\sqrt{2}\] Using Pythagoras theorem                 \[O{{B}^{2}}=O{{C}^{2}}+C{{B}^{2}}\] \[\Rightarrow \]               \[{{(2)}^{2}}=x_{1}^{2}+y_{1}^{2}+{{(\sqrt{2})}^{2}}\] \[\Rightarrow \]               \[x_{1}^{2}+y_{1}^{2}=2\] Hence, locus of mid point of chord is                 \[{{x}^{2}}+{{y}^{2}}=2\]