I am stuck on this complex geometry problem?

This is the question





G be the set of maps f(z)=c*z+d for c,d in the complex plane


Let C1, C2 be circles with centres a1, a2 and radii r1, r2 respectively where a1%26lt;%26gt;a2, r1%26lt;%26gt;r2. Determine


which maps f in G satisfy f (C1) = C2. Show that the fixed points of these maps comprise a circle (called the circle of similitude). [Hint; use Apollonius’ Theorem.] What is the radius of the circle of similitude?





Apollonius' Theorem being that |z − α| = k |z − β| is a circle





I have tried finding the conditions on what c and d must be, but I get slightly different answers each way I do it. Does anyone know how I might go about this?

I am stuck on this complex geometry problem?
Sorry, were the same!





I don't like geometry. It sucks!!!
Reply:I guess the trick is to treat it geometrically, and not get too hung up on the algebra side of it. The map f(z) simply inflates, rotates and translates each point, so pretty easy to convince yourself that it will move C1 to another circle. Moreover it will map circles to circles generally.





Consider |f(z)-ca1-d| = |c(z-a1)|=|c|r1, therefore f(C1) is a circle of radius |c|r1 centred at ca1+d. Hence we must have:





a2=ca1+d and |r2|=|cr1|, (*)





these are the only conditions on c and d.





Now we want the fixed points, so f(z)=cz+d=z, i.e. z=d/(1-c). Note if c=1 we must have d=0, but that would imply a1=a2 and r1=r2, which the question excluded. Hence we may assume 1-c%26lt;%26gt;0.





Not too sure where to go from here. Started delving into different coordinates, but looks messy and unnecessary. Not sure where Apollonius' theorem would fit in.