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The equation of the circle, which is touched by the line y = x, has its centre on the positive direction of the x-axis and cut off a chord of

length 2 units on the line √3 y = x, is


Let the equation of the circle be in from of
diff. w.r.t x
 \frac{dy}{dx}=- \frac{x-a}{y}
at the point where the line y=x let the co-ordinates be (m,m)
therefore slope at y=x is 1
 \frac{dy}{dx}=- \frac{m-a}{m}=1
m= \frac{a}{2}
sub. the point (m,m)=(a/2,a/2) in the equation of the circle
u get
r^2= \frac{a^2}{2}
then the equation of circle will be in form
 x^{2} +y^2-2ax- \frac{a^2}{2}=0
therefore more work on it
u will find that equation is

 (x- \frac{1}{2 \sqrt{2} } )^{2}+ y^{2}= \frac{1}{16}

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