The line y = x cuts the circle x^2 + y^2 - 2x | vedclass

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(A) Given circle is $x^2 + y^2 - 2x = 0$  $(i)$ and the line is $y = x$  $(ii)$.Substituting $y = x$ in $(i)$,we get $x^2 + x^2 - 2x = 0$,which simpli

Vedclass · Accepted Answer

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

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(A) Given circle is $x^2 + y^2 - 2x = 0$  $(i)$ and the line is $y = x$  $(ii)$.Substituting $y = x$ in $(i)$,we get $x^2 + x^2 - 2x = 0$,which simplifies to $2x^2 - 2x = 0$.This gives $2x(x - 1) = 0$,so $x = 0$ or $x = 1$.For $x = 0$,$y = 0$,so $A = (0, 0)$.For $x = 1$,$y = 1$,so $B = (1, 1)$.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 the points $(0, 0)$ and $(1, 1)$,we get $(x - 0)(x - 1) + (y - 0)(y - 1) = 0$.This simplifies to $x(x - 1) + y(y - 1) = 0$,which is $x^2 - x + y^2 - y = 0$ or $x^2 + y^2 - x - y = 0$.

The line $y = x$ cuts the circle $x^2 + y^2 - 2x = 0$ at points $A$ and $B$. Find the equation of the circle having $AB$ as its diameter.

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