Question 200379: Two circles of radius r and 2r overlap (not over the centres) such that
they have a common chord of length 2c (using 2c since easier to "handle").
The measurement across the two circles perpendicular to the chord is k.
(this measurement goes through the circles' centers, of course).
Givens are c and k. What is r in terms of c and k?
This is not homework, but a problem I've made up.
I have worked on this for a while; as example:
equating areas from Heron and the usual base*height/2:
(48k)r^3 - (36c^2 + 44k^2)r^2 + (24c^2k + 12k^3)r - (4c^2k^2 + k^4) = 0.
Tried other ways, like combination of Sine/Cosine laws:
but I keep ending up with an irreducible cubic.
Am I trying for something impossible? Thank you for any help.
Answer by RAY100(1637)   (Show Source):
You can put this solution on YOUR website! A sketch would help, but perhaps you could make one as we go.
.
We understand that two circles of radius, r & 2r intersect. You want to define the length of the common chord thru both centers, k.
.
If the circles just touched, the distance would be the sum of the diameters,
(r+r) + ( 2r + 2r)=6r
.
However as they overlap, this is diminished by the distance from the circle at tangency, to the intersect with the perpendicular chord. If we LET that be "a" for the small circle and "b" for the large circle,,
k= 6r -a -b.
.
Now, to find a & b. Looking at the small circle first, THE LINE is perpendiculat to a chord of length 2c. The 1/2 chord is of course "c" . The triangle created is familiar, one leg "c" ,
one leg (hypotenuse) radius, and third leg unknown,, LET it be "e".
.
Using Pythagorous Theorem, e^2 = R^2 - c^2, or e = sqrt(r^2 -c^2).
.
Going back to the sketch, e + a = r, or e= r-a but also = sqrt (r^2 - c^2),,,,
Therefore a = r - sqrt( r^2 -c^2)
.
Likewise for larger circle (r=2R),,,,,,b= 2r - sqrt( (2r)^2 -c^2)
.
Going back to k = 6r -a -b,,, substitute in and simplify to
.
k = 3R +sqrt( R^2 -c^2) + sqrt( 4R^2 -c^2 )
.
.
.
The answer can now be seen as k= R + 2R + e + d ( legs of triangles ) ok
|
|
|
