What is the length of [math]\overline{P_1P_2}[/math]?

>>8482944
C(Center of Circle)P1:
cos theta1 * r = CP1
CP2:
cos theta 2 * (r + x) = CP2
Law of cosines:
P1P22 = CP12 + CP22 - 2*CP1*CP2*cos (theta 2- theta 1).

>>8482957
P1P2 = CP1 + CP2 - 2*CP1*CP2*cos (theta 2- theta 1).*

Okay so assuming that r is at the middle of the radius and you know the radius:
-go ahead with trig to get the vertical cathetes (rP1 and the other one for P2)
-find the area of the whole figure and the areas of the parallelogram and of that small triangle, subtract the 2 from the big area and you end up with the area of the upper triangle where P1P2 is
-now get this area via sine theorem
And you find OP2
-perpendicular from P1 to OP2 and you find where it lands (you can check that is the middle)
-good old Pythagoras to get P1P2 then
Might have skipped a few steps but this is the outline

>>8482944
First, you need to give an orientation, but who gives a shit with such a shitty diagram you gave. Let's assume you meant P1P2.

Let's call p the length defined for P2 like r for P1.

P1P2 = sqrt ( x^2 + y^2 )
x = p-r
y = p tan(theta2) - r tan(theta1)

>>8483373
>First, you need to give an orientation

Your orientation is fucking homosexual.

>>8483425
>orientation
Yours is fucking inexistent.
So is your intelligence.