The equation of the circle having x-y-2=0 and x-y+2=0 as two langents and x-y=0 as diameter, is

Options

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`

Answer

C

Explanation

Since, the equation of tangents x-y-2=0 and x-y+2=0 are parallel. `therefore` distance between them= diameter of the circle `=(2-(2))/(sqrt(1^(2)+1^(2)))=4/sqrt2=2sqrt2` `[therefore d=(c_(2)-c_1)/(sqrt(a^(2)+b^(2)]]
`radius =1/2(2sqrt2=sqrt2 it is clear from the that centre lies on the origin. `therefore` equation of cricle is `(x-0)^(2)+(y-0)^(2)=(sqrt2)^(2)` rArrx^(2)+y^(2)=2`