Two circles with radii a and b touch each other externally. Let c be the radius of a circle which touches?

Question

these two circles as well as a common tangent to the two circles. Prove that : 1/ (sqr root of c) = 1/(sqr root of a) + 1/ (sqr root of b)

Answer 1

Okay, first draw a line from center of circle c to and perpendicular to the tangent. Call the length of segments of the tangent from circle a and circle b from this point of intersection, x and y. Using the good old Pythagorean theorem, we immediately have the following:

(a+c)^2 - (a-c)^2 = x^2
(b+c)^2 - (b-c)^2 = y^2
(a+b)^2 - (a-b)^2 = (x+y)^2 ....that is, for each:

4ac = x^2
4bc = y^2
4ab = (x+y)^2 ...that is, for each:

2√ac = x
2√bc = y
2√ab = x+y = 2√ac + 2√bc ...division by 2√abc gets you:

1/√c = 1/√b + 1/√a

Answer 2

OMG! I should have paid better attention in Geometry...or is that Trig?

Answer 3

Difficult to answer