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

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(B) We have the equation of the line $y = x$ and the equation of the circle $x^2 + y^2 - 2x = 0$.To find the intersection points of the given line and

Vedclass · Accepted Answer

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

Vedclass · Answer

(B) We have the equation of the line $y = x$ and the equation of the circle $x^2 + y^2 - 2x = 0$.To find the intersection points of the given line and circle,substitute $y = x$ into the circle equation:$x^2 + x^2 - 2x = 0$$2x^2 - 2x = 0$$2x(x - 1) = 0$$x = 0, 1$When $x = 0$,$y = 0$; when $x = 1$,$y = 1$.Thus,the coordinates of the endpoints of the diameter $AB$ are $(0, 0)$ and $(1, 1)$.The equation of a circle with diameter endpoints $(x_1, y_1)$ and $(x_2, y_2)$ is given by $(x - x_1)(x - x_2) + (y - y_1)(y - y_2) = 0$.Substituting the points $(0, 0)$ and $(1, 1)$:$(x - 0)(x - 1) + (y - 0)(y - 1) = 0$$x(x - 1) + y(y - 1) = 0$$x^2 - x + y^2 - y = 0$$x^2 + y^2 - x - y = 0$

Column $I$	Column $II$
$(A)$ Two intersecting circles	$(p)$ have a common tangent
$(B)$ Two mutually external circles	$(q)$ have a common normal
$(C)$ Two circles,one strictly inside the other	$(r)$ do not have a common tangent
$(D)$ Two branches of a hyperbola	$(s)$ do not have a common normal

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 a diameter is . . . . . .

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