A) \[{{x}^{2}}+{{y}^{2}}+2x-2y+1=0\]
B) \[{{x}^{2}}+{{y}^{2}}-2x+2y-1=0\]
C) \[{{x}^{2}}+{{y}^{2}}=2\]
D) \[{{x}^{2}}+{{y}^{2}}=1\]
Correct Answer: C
Solution :
Since, the equations of tangents\[x-y-2=0\] and\[x-y+2=0\]are parallel. \[\therefore \]Distance between them = diameter of the circle \[=\frac{2-(-2)}{\sqrt{{{1}^{2}}+{{1}^{2}}}}\] \[\left( \because \frac{{{C}_{2}}-{{C}_{1}}}{\sqrt{{{a}^{2}}+{{b}^{2}}}} \right)\] \[=\frac{4}{\sqrt{2}}=2\sqrt{2}\] \[\therefore \]Radius \[=\frac{1}{2}(2\sqrt{2})=\sqrt{2}\] It is clear from the figure that centre lies on the origin. \[\therefore \]Equation of circle is \[{{(x-0)}^{2}}+{{(y-0)}^{2}}={{(\sqrt{2})}^{2}}\] \[\Rightarrow \] \[{{x}^{2}}+{{y}^{2}}=2\]You need to login to perform this action.
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