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given 3 circles, the circumference of the 2nd circle cuts the centre of circle1 at O1, while the centre of the 3rd circle is formed between the upper intersection of 1st and 2nd circle, with its circumference cuts through centre of 1st circle. radius of 2nd circle is unknown.
and a str8 line is formed between O1 and O2, with P be the right intersection of 3rd circle with the line.
O1-P to O1-O2 are these distances
o1 to p i the base of an iso tri with sides of r|dw:1344615311196:dw|
we can define an angle using o1,o2 and oi,1 so that o1,p is findable to me
err ... o1,o3 that is
or really P = (2a,0) in this case :)
|dw:1344615575595:dw| im sure we could come up with something
we can determine a general ratio; and when given specifics we can fill them in
so i will need to come out with 3 ratios for O1P < O1O2, O1P = O1O2, and O1P > O1O2?
because radius for circle 2 is not fixed
radius for circle 2 is "fixed"
owh. k i'll try from here. thx =)
i might be reading the post wrong tho
o2 is solid, so do we assume we know the position of o2 relative to the other solid objects?
erm o2 is the centre of 2nd circle. lol i drawed the dot too big i guess.
its an imaginary centre because radius 2 is unknown.
but centre of circle 2 will be placed same level with circle 1
|dw:1344616423539:dw| well, you do have 3 points of contact for the circle2 i believe
the half the length from 03 to 0'3 would be known (y) and half the distance from 01 to P can be known (x) y^2 = x*(distance from p/2 to 03)
|dw:1344616835067:dw| y^2 = x*c ; x+c = radius from 01 to 02
or rather r' = x+c
the ratio then is.... 2x : r'
2x : x+c , and since y^2 = xc; c=y^2/x 2x : x + y^2/x 2x : (x^2 + y^2)/x
o i'll try to digest 'em. really thx for spending time with this question. @@
youre welcome, it was fun :)
y^2 = xc is some theorem? o.o