If four distinct points (2k, 3k), (2,0), (0,3 | vedclass

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(C) The points $(2,0)$, $(0,3)$, and $(0,0)$ form a right-angled triangle at the origin $(0,0)$ because the angle between the $x$-axis and $y$-axis is

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$k = 1$

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(C) The points $(2,0)$, $(0,3)$, and $(0,0)$ form a right-angled triangle at the origin $(0,0)$ because the angle between the $x$-axis and $y$-axis is $90^{\circ}$.Since the angle subtended by the line segment joining $(2,0)$ and $(0,3)$ at the point $(0,0)$ is $90^{\circ}$, this line segment must be the diameter of the circle.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 $(2,0)$ and $(0,3)$, we get:$(x-2)(x-0) + (y-0)(y-3) = 0$$x^2 - 2x + y^2 - 3y = 0$$x^2 + y^2 - 2x - 3y = 0$Since the point $(2k, 3k)$ lies on this circle, it must satisfy the equation:$(2k)^2 + (3k)^2 - 2(2k) - 3(3k) = 0$$4k^2 + 9k^2 - 4k - 9k = 0$$13k^2 - 13k = 0$$13k(k - 1) = 0$This gives $k = 0$ or $k = 1$.If $k = 0$, the point $(2k, 3k)$ becomes $(0,0)$, which is not a distinct point from the given point $(0,0)$.Therefore, for the four points to be distinct, we must have $k = 1$.

If four distinct points $(2k, 3k)$, $(2,0)$, $(0,3)$, and $(0,0)$ lie on a circle, then:

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