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In this truly amazing answer to a question I had asked this morning, a following construction is done:

Consider the a vector from origin to a point on the $(x,y) $ plane, also a circle $C$ having center $(a,b)$. The vector from the circle $C$ center to this point is $(x-a,y-b)$ , now it seems that the set of points for which these two vectors are perpendicular are another circle:

$$ (x,y) \cdot (x-a,y-b) = 0$$

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

But I just can't understand why this set of points satisfying above two criteria must be a circle, could someone shed some intuition on why we should expect this? You can find an example of the above discussion here

Clemens Bartholdy
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1 Answers1

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The locus of points subtending a right angle on the segment $OC$ is a circle with diameter $OC$ and center its midpoint.

The above equation can be rewritten more clearly as $$\left(x-\frac{a}{2} \right)^2+\left(y-\frac{b}{2} \right)^2=\left(\frac{\sqrt{a^2+b^2}}{2} \right)^2=\left(\frac{OC}{2} \right)^2$$

cosmo5
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