Thread: calculating with circles View Single Post
 2014-07-12, 12:14 #10 diep     Sep 2006 The Netherlands 2·73 Posts hi wblipp, that is indeed one of the 2 slopes. As for the math it seems that there is another manner to calculate it. If we take the length from the r=1.00 circle to the r=0.555 circle, and we already know that its position is (x,y) = (-0.4,0.195) Then the distance of that i noticed is sqrt( 0.195^2 + 0.4^2 ) = 0.445 Which happens to be 1 - 0.555 So the line that defines the point where the circles touch goes from origin of the 1.00 circle through the origin of the 0.555 circle In short we can then easily calculate its position as distance from origin of the 0.555 circle: (555/445) * {x = 0.4 ,y = -0.195} = {0.498876404 , -0.243202247} If i fill in then into: x^2 + y^2 = 0.555^2 ==> y^2 = 0.555^2 - 0.498876404^2 ==> y = +- 0.243202247 Now let's fill in in the formula : (x +- 0.4) ^2 + (y - 0.195) ^2 = 1 <==> y^2 - 0.39y + 0.195^2 - 1 + (x +- 0.4) ^2 = 0 Then let's calculate for both +0.4 as well as -0.4 ==> (edit: with better numbers...) A : y^2 - 0.39y + 0.195^2 - 1 + 0.807978791 = y^2 - 0.39y - 0.153996209.. B : y^2 - 0.39y + 0.195^2 - 1 + 0.009776543 = y^2 - 0.39y - 0.952198457.. D = b^2 + 4ac ==> A: D = 0.768084838.. B: D = 3.960893827.. A: x1,x2 = (0.39 +- sqrt(D)) / 2 = (0.39 +- 0.876404494.. )/2 = { -0.243202247 , 0.633202247 } one solution outside the 0.555 circle B: x1,x2 = (0.39 +- sqrt(D)) / 2 = (0.39 +- 1.990199444...)/2 = { -1.600199444 / 2 , 2.380199444 / 2} both solutions outside the 0.555 circle So that's the answer i was looking for (after the edit). Last fiddled with by diep on 2014-07-12 at 12:29