Given a circle and a point K inside it. An arbitrary circle equal to the given one and passing through the point K has a common chord with the given circle. Find the geometric locus of the midpoints of these chords.
Let the given circle be centered at A=(-2a,0) and have radius r.
Let K=(0,0)
The arbitrary circle has a center lying on the circle centered at point C=(c,d) with c^2+d^2=r^2 [a constant]
The two circles intersect at B and D and ABCD is a rhombus. Thus the midpoint of the chord BD is also the midpoint of chord AC=E=(-a+c/2,d/2).
Since we suspect the locus of point E is a circle centered at the midpoint of AK=F=(-a,0) let's find the distance EF
distance^2 = (c/2)^2+(d/2)^2 = (c^2 + d^2)/4 = r^2/4
distance = r/2
This shows that not only is this distance a constant this constant is half of r.
In retrospect, I see the simple geometric proof.
ABCD is a rhombus with side length r, and E is the midpoint of AC, we have triangle AKC with KC=r. Since E and F are midpoints of sides AC and AK respectively, EF=KC/2=r/2.
So the locus is a circle with radius r/2 centered at the midpoint of AK
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Posted by Jer
on 2021-05-01 11:30:22 |
two proofs
