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 \[{{x}^{2}}+{{y}^{2}}=2\]You need to login to perform this action.
You will be redirected in
3 sec