Circle and Tangent Line

Let $C$ be the circle with radius $r$, $x^2 + y^2 = r^2$. We choose two points $P(a, b)$ and $Q(-a, c)$ so that the line passing through $P$ and $Q$ is tangent to $C$.
For example, the quadruplet $(r, a, b, c) = (2, 6, 2, -7)$ satisfies this property.
Let $F(R, X)$ be the number of the integer quadruplets $(r, a, b, c)$ with this property, and with $0 \lt r \leq R$ and $0 \lt a \leq X$.
We can verify that $F(1, 5) = 10$, $F(2, 10) = 52$ and $F(10, 100) = 3384$.
Find $F(10^8, 10^9) + F(10^9, 10^8)$.

This problem seems to be a part of a competitive math problem set, especially that it does not contain a typical curriculum concept.

This problem is complex because it involves circle geometry, tangents, points, line equations, and number theory to find the sum of two functions $F(10^8, 10^9) + F(10^9, 10^8)$.

Let’s find out how we can solve a simpler version of the problem and the approach for solving it, so you can get the main idea.

1) First, you need to understand that a line that is tangent to a circle at a certain point will have the same slope as the radius to that point. That’s because a radius perpendicular to a tangent line curves at the point of tangency.

2) From the equation of the circle $x^2 + y^2 = r^2$, we can see that it is centered at the origin (0,0) with a radius r.

3) If we select two points P(a, b) and Q(-a, c) that a tangent line passes through, we can say that the midpoint M of line PQ lies on y=0 line because, on adding the y-coordinates and dividing by 2 [(b+c)/2], we get 0.

4) Now calculate the slope of MQ, it will be (c-b)/2a.

5) This slope should be equal to the slope of the radius from the center of the circle to the point of tangency. Considering the circle equation, the derivative (slope of the radius) would be -x/y.

6) Equating the two slopes and solving the equation will bring you to some relationship that are integer solutions a, b, and c for fixed r, which means that for a fixed radius, many quadruplets (r, a, b, c) could satisfy the provided conditions.

7) For each radius r from 1 to R, count the number of integer solutions for a from 1 to X. This will be the answer to $F(R, X)$.

Note: This approach outlines the method to solve this kind of problem but doesn’t yet provide the direct answer to the problem, as determining the specific number of quadruplets with such large values of R and X ($F(10^8, 10^9) + F(10^9, 10^8)$) would need a lot of computational power and possibly application of special algorithms for fast calculations.

It’s also worth mentioning that this problem mixes geometry with number theory in a nontrivial way, so it may be designed as a challenge problem for high-level math competitions or advanced study rather than a routine homework problem or test question.

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