Problem about triangle 1

See diagram: http://img136.imageshack.us/img136/2826/circlesn.jpg
Since ∠APC = 45 + 120 = 165, P must lie above line AC in the diagram.
Let the points be fixed in a co-ordinate system such that B is (0,0) and A is (-2,0).
Any point T with ∠ATB = 45 has a locus of an arc of a circle C1 as shown. P is on this arc. Since the inscribed angle of the arc is 45, the inscribed angle to the centre of the cicrle is 45 x 2 = 90 degree.
With OA = OB = radius, OA2 + OB2 = 22 => radius = √2
Let the co-ordinate of O be (c,d)
(c + 2)2 + (d – 0)2 = 2
(c – 0)2 + (d – 0)2 = 2
So c = 1 and d = 1.
Equation of C1 is (x + 1)2 + (y – 1)2 = 2
or x2 + 2x + y2 – 2y = 0 … (1)
Similarly, any point S with ∠BPC = 120 has a locus of an arc to the circle C2. P is also on the arc. Since ∠BPC = 120, the bigger angle ∠BQC = 120 x 2 = 240. The smaller angle ∠BQC = 360 – 240 = 120.
Since QB = QC, ∠QBC = (180 – 120) / 2 = 30
Since ∠ABC = 150, so Q lies on the x-axis.
By cosine rule, BC2 = QB2 + QC2 – 2(QB)(QC)cos120
3 = r2 + r2 – 2r2(-0.5) where r is the radius of C2
3 = r2 + r2 + r2 => r = 1
Equation of C2 is (x – 1)2 – (y – 0)2 = 1
or x2 – 2x + y2 = 0 … (2)
(1)– (2) => 4x – 2y = 0 or y = 2x
Sub into (2) => x2 – 2x + 4x2 = 0
x(5x – 2) =0
x = 0 (point B) or x = 2/5 (point P)
with x = 2/5, y = 4/5
BP2 = (2/5)2 + (4/5)2
BP = 2√5/5