The points A(a, 0), B(0, b), C(c, 0) and D(0, | vedclass

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(D) The equation of a circle passing through three points $(a, 0), (c, 0)$ and $(0, b)$ can be determined by the power of a point theorem or by findin

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are concyclic

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(D) The equation of a circle passing through three points $(a, 0), (c, 0)$ and $(0, b)$ can be determined by the power of a point theorem or by finding the circumcircle.For points $(a, 0), (c, 0)$ on the $x$-axis and $(0, b), (0, d)$ on the $y$-axis,the points are concyclic if the product of the intercepts on the $x$-axis equals the product of the intercepts on the $y$-axis.That is,$a 	imes c = b 	imes d$.Since the problem states $ac = bd$,the four points $A, B, C,$ and $D$ satisfy the condition for being concyclic.Therefore,the points are concyclic.

The points $A(a, 0), B(0, b), C(c, 0)$ and $D(0, d)$ are such that $ac = bd$ and $a, b, c, d$ are all non-zero. Then the points:

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