Three externally touching circles have their centers on the same line and have radii $a$,$b$ and $c$ (where $aThree circles have their centers on the same line and have radii $a$,$b$ and $c$ (where $a<b<c$).The circle with radius $b$ touches the other two circles but circles with radii $a$ and $c$ do not touch each other.The three circles also have a common tangent.Prove that $b^2=ac$.

I solved it with good deal of calculations. Is there an elementary way to do this problem

Say common tangent length between circle A and B (IJ) is $x$, B and C (JK) is $y$ and so between A and C (IK) is $(x + y)$.

Applying Pythagoras, between circles of radius a and b,
$x^2 = (a+b)^2 - (b-a)^2 = 4ab$
Applying Pythagoras, between circles of radius b and c,
$y^2 = (b+c)^2 - (c-b)^2 = 4bc$
Applying Pythagoras, between circles of radius a and c,
$(x+y)^2 = (a+2b+c)^2 - (c-a)^2 = 4(b^2+ab+bc+ca)$
i.e $x^2 + y^2 + 2xy = 4b^2+4ca+x^2+y^2$
i.e $4b^2+4ca = 2xy = 8b\sqrt{ac}$
i.e $b^2 - 2b\sqrt{ac} + ca = 0$
i.e $b^2 = ca$
