question_answer

    The equation of the circle having\[x-y-2=0\]and\[x-y+2=0\]as two tangents and\[x-y=0\]as a diameter is

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                 \[{{(x-0)}^{2}}+{{(y-0)}^{2}}={{(\sqrt{2})}^{2}}\] \[\Rightarrow \]               \[{{x}^{2}}+{{y}^{2}}=2\]