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# Intercept on the line $y=x$ by the circle $x^2+y^2-2x=0$ is $AB$ equation of the circle on AB as a diameter is :

$\begin{array}{1 1}(A)\;x^2+y^2+x-y-=0 \\(B)\;x^2+y^2-x+y=0 \\(C)\;x^2+y^2+x+y=0 \\(D)\;x^2+y^2-x-y=0 \end{array}$

$y=x$
$x^2+y^2-2x=0$
$x^2+x^2-2x=0$
$2x^2-2x=0$
$2x(x^2-1)=0$
=> $x=0 \;or \;x=1$
=> $y=0 \;or\;y=1$
$\therefore$ extremities of diameter of the required circle are $(0,0)$ and $(1,1)$ Hence the equation of the circle is
$(x-0)(x-1)+(y-0)(y-1)=0$
=>$x^2+y^2-x-y=0$
Hence D is the correct answer.