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overlapping circles with diameters = tangents?
If we are given:
circles O1 and O2 which meet exactly at the points {A,D} so that the diameter AB of O1 is tangential to O2 at the point A, and the diameter AC of O2 is tangential to O1 at the point A (and so AB is perpendicular to AC),
then
the question is: does D lie on the line BC?
To find out, I used A as the origin of a Cartesian coordinate system, with the diameters lining up with the positive x and y axes respectively.
The equation of the line AC is immediate.
The approach of finding the points of intersection of the two circles (in order to see if it lay on the line) was not feasible.
Another approach was to find the points of intersection of L with each of the circles, and then see if the answers coincided.
If I did not make a mistake, then the answers did not coincide unless the diameters were equal.
This is frustrating, as the gut feeling, bolstered by various drawings of various diameters always appearing to show that D lies on AC.
So, 2 questions:
(a) is this result known?
(b) is there an easier way to do this?
(c) am I wrong, and somehow D does lie on AC?
1 Answer
- Dragon.JadeLv 78 years agoFavorite Answer
Hello,
The question is interesting so I will not begrudge your lack of "Hello", "Please" and "Thanks".
We shall use a very well-known property of the circle:
→ Any point P of a circle associated with any of its diameters will describle a right triangle whose right angle is at P.
Applied to your problem:
D is a point of circle O₁ with diameter AB. So angle BDA is right.
D is also a point of circle O₂ with diameter AC. So angle CDA is right.
Now we have two cases:
→ angle BDC = BDA + ADC = 90° + 90° = 180° → and points B, D and C are aligned.
→ angle BDC = BDA – ADC = 90° – 90° = 0° → and points B, D and C are aligned.
So your hunch is correct, D is always on line BC.
NOTE:
Notice that the fact that both circles being tangent to the diameter of the other has no bearing on the reasoning above.
So this property works even if the diameters are not at right angle.
Regards,
Dragon.Jade :-)