Question

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

Answer 1

let the equation of the circle be in from of
$(x-a)^2+y^2=r^2$
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
$\frac{dy}{dx}=- \frac{m-a}{m}=1$
solve
$m= \frac{a}{2}$
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
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$(x- \frac{1}{2 \sqrt{2} } )^{2}+ y^{2}= \frac{1}{16}$