Bmo1 q3 2011/2012

I've got what I hope is a solution but I've never used any techniques like this before so would really appreciate feedback.

Proof:

Let the centres of S, S1, S2 be O,O1,O2
I drew the tangent to S and S2 as I pictured them to be internally tangent and called it L1.
OYL1=O2YL2=90 so O,O2,Y are collinear. Since the line of tangency and line of centers intersect at Y, Y is a centre of similtude and since PXY goes through Y OX is parallel to O2P. Also O1,X,O are collinear. Thus angles YOX=XO1P=YO2P. Denote the radii of circles like that of centres and by cosine rule: XY^2=2r^2-2r^2costheta and then do the same for PX and PY..... aafter tedious working I got to r=lr2-r1l. Could anyone tell me how many marks I would've gotten. Many thnx

Original post by 11234

I've got what I hope is a solution but I've never used any techniques like this before so would really appreciate feedback.

Proof:

Let the centres of S, S1, S2 be O,O1,O2
I drew the tangent to S and S2 as I pictured them to be internally tangent and called it L1.
OYL1=O2YL2=90 so O,O2,Y are collinear. Since the line of tangency and line of centers intersect at Y, Y is a centre of similtude and since PXY goes through Y OX is parallel to O2P. Also O1,X,O are collinear. Thus angles YOX=XO1P=YO2P. Denote the radii of circles like that of centres and by cosine rule: XY^2=2r^2-2r^2costheta and then do the same for PX and PY..... aafter tedious working I got to r=lr2-r1l. Could anyone tell me how many marks I would've gotten. Many thnx