(B) To find the points of intersection $A$ and $B$, substitute $y=x$ into the circle equation $x^2+y^2-2x=0$.$x^2+x^2-2x=0$$2x^2-2x=0$$2x(x-1)=0$Thus,

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(B) To find the points of intersection $A$ and $B$, substitute $y=x$ into the circle equation $x^2+y^2-2x=0$.$x^2+x^2-2x=0$$2x^2-2x=0$$2x(x-1)=0$Thus, $x=0$ 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 $(0,0)$ and $(1,1)$:$(x-0)(x-1)+(y-0)(y-1)=0$$x(x-1)+y(y-1)=0$$x^2-x+y^2-y=0$$x^2+y^2-x-y=0$.

Class 11 Mathematics 10-1.Circle and System of Circles Geometrical problems regarding circle and its properties

Medium

The intercept on the line $y=x$ by the circle $x^2+y^2-2x=0$ is $AB$. The equation of the circle with $AB$ as diameter is:

WBJEE 2011 All WBJEE

A
$x^2+y^2=1$
B
$x^2+y^2-x-y=0$
C
$x^2+y^2=2$
D
$x^2+y^2-2x-2y+1=0$

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10-1.Circle and System of Circles

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Geometrical problems regarding circle and its properties

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The intercept on the line $y=x$ by the circle $x^2+y^2-2x=0$ is $AB$. The equation of the circle with $AB$ as diameter is:

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