let the equation of the circle be in from of
[tex](x-a)^2+y^2=r^2[/tex]
diff. w.r.t x
[tex] \frac{dy}{dx}=- \frac{x-a}{y} [/tex]
at the point where the line y=x let the co-ordinates be (m,m)
therefore slope at y=x is 1
[tex] \frac{dy}{dx}=- \frac{m-a}{m}=1 [/tex]
solve
[tex]m= \frac{a}{2} [/tex]
sub. the point (m,m)=(a/2,a/2) in the equation of the circle
u get
[tex]r^2= \frac{a^2}{2} [/tex]
then the equation of circle will be in form
[tex] x^{2} +y^2-2ax- \frac{a^2}{2}=0 [/tex]
therefore more work on it
u will find that equation is
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[tex] (x- \frac{1}{2 \sqrt{2} } )^{2}+ y^{2}= \frac{1}{16} [/tex]
