Prove that the equation of the circle described on the line joining (a, b) and (c, d) as diameter is (x-a)(x-c)+(y-b)(y-d) = 0.
Source: G. H. Hardy, A Course of Pure Mathematics, Rough Draft Printing, 2007.
Solution 1 (due to Han Qing).
Let A = (a, b), B = (c, d), and P = (x, y) be an arbitrary point on the given circle. Consider the vectors AP = (x-a, y-b) and BP = (x-c, y-d). Since APB is an angle subtended by the diameter AB of the circle, APB is a right angle. It follows that AP and BP are perpendicular, and their dot product is 0. Hence (x-a)(x-c)+(y-b)(y-d) = 0.
Since (a, b) and (c, d) are the endpoints of the diameter of the given circle, ((a+c)/2, (b+d)/2) is the center of the circle, and (a-c)^2+(b-d)^2 is 4 times the square of the diameter of the circle. It follows that the equation of the circle is
[x-(a+c)/2]^2+[y-((b+d)/2]^2 = [(a-c)^2+(b-d)^2]/4,
x^2-ax-cx+y^2-by-dy = [(a-c)^2-(a+c)^2+(b-d)^2-(b+d)^2]/4 = -ac-bd,
(x-a)(x-c)+(y-b)(y-d) = -ac-bd+ac+bd = 0.
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