The intercept on the line y = x by the circle | vedclass

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(D) Given,the circle is ${x^2} + {y^2} - 2x = 0$ $(i)$ and the line is $y = x$ $(ii)$.Substituting $y = x$ in $(i)$,we get:${x^2} + {x^2} - 2x = 0$$2{

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${x^2} + {y^2} - x - y = 0$

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(D) Given,the circle is ${x^2} + {y^2} - 2x = 0$ $(i)$ and the line is $y = x$ $(ii)$.Substituting $y = x$ in $(i)$,we get:${x^2} + {x^2} - 2x = 0$$2{x^2} - 2x = 0$$2x(x - 1) = 0$So,$x = 0$ or $x = 1$.Since $y = x$,the points of intersection 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$$x(x - 1) + y(y - 1) = 0$${x^2} - x + {y^2} - y = 0$${x^2} + {y^2} - x - y = 0$.

The intercept on the line $y = x$ by the circle ${x^2} + {y^2} - 2x = 0$ is $AB$. The equation of the circle having $AB$ as a diameter is

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