A Maths (Circles)

(1) Observe that the angle bisectors of L1 and L2 are the x-axis and y-axis, therefore the centre must lies on either one of them. Also observe that point A is lies under y=x, thus this centre only can lie on the x-axis.

Let C: (x-h)^2+y^2=r^2 where h,r>0
distance between centre and y=x is given by
|h-0|/√2=r, i.e. r^2=h^2/2

Putting A(4,2√2) and r^2=h^2/2 into C,
(4-h)^2+(2√2)^2=h^2/2
h^2-16h+48=0, (h-4)(h-12)=0, h=4 or h=12

When h=4, r^2=8 and when h=12, r^2=72
thus the required circles are
(x-4)^2+y^2=8 and (x-12)^2+y^2=72

(2) Let C: (x-h)^2+(y-k)^2=r^2, G=(h,k)

Let M be the mid-point of AB, then AM=BM=1
also ΔGAM is right-angle triangle, thus by Pyth. Theorem,
AM^2+GM^2=GA^2, i.e. 1+k^2=r^2.

Let R be the mid-point of PQ
Since major arc PQ: minor arc PQ=3:1
minor arc PQ is 1/4 of the whole circle,
and therefore ΔGPR is right-angle triangle, also GP=GQ=r,
thus we have GR=h=r/√2, i.e. r=√2*h

Since distance from G to x-2y=0 is d=1/√5
thus d=|h-2k|/√5=1/√5, i.e. |h-2k|=1

Solving them, we get (h,k)=(1,1), (-1,-1), (5/7,-1/7), (-5/7,1/7)
for the first two, r^2=2, for the last two, r^2=50/49

thus the required circles are
(x-1)^2+(y-1)^2=2
(x+1)^2+(y+1)^2=2
(7x-5)^2+(7y+1)^2=50
(7x+5)^2+(7y-1)^2=50

2008-11-15 23:36:48 補充：
Since in the first time, my solution is too long and not allowed to submit, therefore I shorten them into this new version. Hope you can understand.