The 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

First, find the points of intersection $A$ and $B$ by substituting $y=x$ into the circle equation $x^2+y^2-2x=0$:
$x^2+x^2-2x=0$
$2x^2-2x=0$
$2x(x-1)=0\Rightarrow x=0$ or $x=1$.
Since $y=x,$ the points are $A(0,0)$ and $B(1,1)$.The equation of a circle with diameter endpoints $(x_1,y_1)$ and $(x_2,y_2)$ is:$(x-x_1)(x-x_2)+(y-y_1)(y-y_2)=0$.Substituting $A(0,0)$ and $B(1,1)$:$(x-0)(x-1)+(y-0)(y-1)=0$$x^2-x+y^2-y=0$$x^2+y^2-x-y=0$.