In the figure shown find the relation between $a,b,c$.
My try: When two circles of radii $r_1,r_2$ touch externally, the length of their direct common tangent is $2\sqrt{r_1r_2}$ Let the radius of the bigger circle is $r_1$ and that of smaller one is $r_2$. We have:
$AM=a$
$MK=2\sqrt{r_1r_2}$
By Pythagoras theorem we get: $$AK^2+4r_1^2=(a+b+c)^2$$
Also by secant tangent theorem we have: $$AK^2=AC \times AD$$ So we get: $$AK^2=(a+b)(a+c)$$ So we get: $$4r_1^2=b^2+c^2+ab+bc+ca$$
Any hint from here?
If you let angle KAD = 2p, then sin(2p) = 2(r1)/(a+b+c) from triangle AKD
from a circle theorem angle KCD is a right angle and angle CKD=2p, so c=2(r1)sin(2p) and so c=4r(1)^2/(a+b+c)
then perhaps you can use what you did to eliminate r1
...after trying it, it leads to an inconsistency, there might be a mistake in your AK^2 line, second bracket should be (a+b+c), then it's the same result...would then recommend using triangle AB(O2), where O2 is the centre of circle 2, tan(p)=(r2)/a and cos(2p)=AK/(a+b+c) from triangle AKD and then use trig identities to try and eliminate unwanted letters.
...further results:
$$4r_1^2=c(a+b+c) $$
$$AK = \sqrt{(a+b)(a+b+c)}$$
$$sin(2p)=\sqrt{\frac{c}{a+b+c}}$$
$$cos(2p)=\sqrt{\frac{a+b}{a+b+c}}$$
$$tan(2p)=\sqrt{\frac{c}{a+b}}$$
so perhaps they can be related using inverse trig functions, all equalling 2p and each other.