How to get radius of lower and upper bound of circle? - algorithm

Question

Problem and nearby solution image:
Problem: A mesh equally distant from each node. Each small node of grid is a circle(fig: grey circle) of radius r, if we make any circle(fig:blue color) then how can we make a set of that circle having upper(fig:red circle) and lower bound(fig:red circle). Basically we need to find upper bound radius, and lower bound radius.
Value known are:
Radius of small circle(grey) r.
Distance between two grey circle/node_of_grid from center to center = 4 X r.
Radius of big circle (blue) rb. //this may vary but will be in multiple of r. In this figure case is diameter d = 2(20 x r).
Center of big circle can only be center of small circle.
How can i find the radius of upper circle and lower circle. Each blue small circle who comes within the big circle circumference should come under upper bound and vice versa for lower bound.
Currently i came up with this solution which is incorrect :
unpper_bound_radius = rb + (r + (rb/(3.1415926 * 2r)))
lower_bound_radius = rb - (r + (rb/(3.1415926 * 2r)))
Thanks

Answer 1

If you're looking for an algorithm to decide which grey dots intersect with the blue circle, and calculate the radius of the boundaries that contain the selected grey dots, you could do something like this:
Give the inner and outer boundary a radius equal to the blue circle.
The grey dot straight above the center, which intersect with the blue circle or is just above it, is the starting point.
Calculate the distance from that point to the center point; if it is between rb-r and rb+r, the blue circle intersects with this grey dot.
If it intersects, adjust the inner boundary to distance - r and the outer boundary to distance + r as necessary.
If it doesn't, and the distance is greater, try the grid point below it; if it doesn't, and the distance is smaller, try the grid point to the right.
Do this until you move out of a 45° sector; you only need to check one eigth of the circle, the other results will be symmetrical.
Example calculation:
The first grid point, at the top, is 5×4r from the center; the boundaries are at 5×4r-r and 5×4r+r.
The next grid point to the right is at 5.09902×4r from the center, so the outer boundary increases to 5.09902×4r+r.
The next grid point to the right is at 5.38516×4r from the center, which is 1.54066×r outside the blue circle (too far).
The next grid point downward is at 4.47216×4r from the center, which is 1.88854×r inside the blue circle (too close).
The next grid point to the right is at exactly 5×4r from the center, so the boundaries don't change.
The next grid point to the right is at 5.65686×4r from the center, which is 2.62742×r outside the blue circle (too far).
The next grid point down is more than 45° to the right.
Result: inner boundary is at 5×4r-r, outer boundary is at 5.09902×4r+r.
function boundaries(b) {
var i = b, o = b, x = 0, y = Math.ceil(b / 4) * 4;
while (Math.atan2(y, x) >= Math.PI / 4) {
var d = Math.sqrt(x * x + y * y);
if (d < b - 1) x += 4; // too close, go right
else if (d > b + 1) y -= 4; // too far, go down
else { // intersection
if (i > d - 1) i = d - 1; // adjust inner
if (o < d + 1) o = d + 1; // adjust outer
x += 4; // go right
}
}
return {i: i, o: o};
}
var res = boundaries(20);
document.write("inner: r×" + res.i + "<br>outer: r×" + res.o);