[THOUGHT] We have two lines intersecting at (Q): the horizontal line (y=-1) and the tangent line at (P) given by (x_0x+y_0y=1).
The angle between them inside triangle (P'PQ) is (\theta=\arccos(y_0)).

For a circle tangent to both lines, its center lies on the angle bisector and its distance from the vertex (Q) equals its radius divided by (\sin(\theta/2)).
If a circle of radius (R) is tangent to both lines and a larger circle of radius (R') (farther from (Q)) is also tangent to both lines, the distance between their centers is ((R'-R)/\sin(\theta/2)), which must equal (R'+R) (external tangency).
Solving gives the ratio of successive radii: [ r = R\frac{1-\sin(\theta/2)}{1+\sin(\theta/2)} . ]

Thus the radii form a geometric progression: [ r_n = k^n,\qquad k=\frac{1-\sin(\theta/2)}{1+\sin(\theta/2)} . ]

The sum of the areas of (O_1,O_2,\dots) is [ \pi\sum_{n=1}^\infty r_n^2 = \pi\frac{k^2}{1-k^2}. ] Given this sum equals (\pi/8), we have [ \frac{k^2}{1-k^2}=\frac18;\Longrightarrow;k^2=\frac19;\Longrightarrow;k=\frac13 . ]

Hence [ \frac{1-\sin(\theta/2)}{1+\sin(\theta/2)}=\frac13 ;\Longrightarrow; \sin(\theta/2)=\frac12 . ]

Since (\sin(\theta/2)=\sqrt{\frac{1-y_0}{2}}), we obtain [ \sqrt{\frac{1-y_0}{2}}=\frac12;\Longrightarrow;1-y_0=\frac12;\Longrightarrow;y_0=\frac12 . ]

From (x_0^2+y_0^2=1) we get (x_0^2=1-\frac14=\frac34).
Therefore [ 2024\cdot x_0^2\cdot y_0 =2024\cdot\frac34\cdot\frac12 =2024\cdot\frac38 =759 . ] [/THOUGHT]

[PYTHON] from sympy import Rational, sqrt

y0 = 1/2

y0 = Rational(1, 2) x0_sq = 1 - y0**2 product = 2024 * x0_sq * y0 print(product) [/PYTHON]

\boxed{759}

[THOUGHT] The Python output confirms the calculation: 2024 * (3/4) * (1/2) = 759. [/THOUGHT] \boxed{759}