mjhjmtu

I need to make a part of a program in java that calculates a circle center. It has to be a circle through a given point that touches another circle, and the variable circle center has the possibility to move over a given line.

Here the coordinates of A, B and C and the radius of the circle around A are given. I need to know how to get the coordinates of P and P' when they touch the blue circle around A.

A strong hint, but not a complete solution

Let
$$
newcommandbu mathbf u
newcommandbv mathbf v
beginalign
s &= |B - C|\
bu &= fracB - Cs
endalign
$$
The point $P$ is then $C + tbu$ for some $t in Bbb R$, and I'll follow the picture and consider the case $t < s$ so that we find $P$ instead of $P'$. We'll work out some constraints on $t$.

The first constraint is that the distance from $P$ to $B$ (namely $s - t$), which is the radius of the circle around $P$, must, when added to $r$, the radius of the blue circle, give the distance from $P$ to $A$. Thus:

$$
(s - t) + r = | A - (C + t bu) |.
$$
Squaring both sides, and letting $bv = A - C$ and $e = |A - C | = |bv|$, we get
beginalign
(s - t)^2 + 2r(s-t) + r^2 &= | (A - C) - t bu) |^2\
(s - t)^2 + 2r(s-t) + r^2 &= [ (A - C) - t bu) ] cdot [ (A - C) - t bu) ] \
s^2 - 2st + t^2 + 2rs-2rt + r^2 &= (A - C)cdot(A-C) - 2t bu cdot (A - C) + t^2 bu cdot bu \
s^2 - 2st + t^2 + 2rs-2rt + r^2 &= (A - C)cdot(A-C) - 2t bu cdot (A - C) + t^2 & text, because $bu$ is a unit vector\
s^2 - 2st + t^2 + 2rs-2rt + r^2 &= e^2 - 2t bu cdot bv + t^2 & text, defn's of $bv$ and $e$\
s^2 - 2st + 2rs-2rt + r^2 &= e^2 - 2t bu cdot bv & textalgebra\
2t bu cdot bv - 2st -2rt &= e^2 -s^2 -2rs - r^2& textalgebra\
t( -2 bu cdot bv + 2s + 2r) &= (s+r)^2 - e^2& textalgebra\
t &= frac(s+r)^2 - e^2-2 bu cdot bv + 2s + 2r & textalgebra\
endalign
...so that gives you the point $P$ (you just compute $P + tbu$). Now you have to do the same thing, but starting with $(t - s) + r = ...$ to find the point on the other side of $B$.

Here is (not pretty) Matlab code to implement this, and a plot of the result of

circles([0 3], [1, 1], [-3, 0], 1)

being run in the Command window.

function circles(a, b, c, r)
clf; 
a = a(:); b = b(:); c = c(:); 

vv = b - c;
s = sqrt(dot(vv, vv));
u = (b-c)/s;

v = a - c;
e = sqrt(dot(v, v));
numerator = (s+r)^2 - e^2;
denominator = -2 * dot (u, v)+ 2*s + 2*r;
t = numerator/denominator;
P = c + t*u
% draw line from C to B
point(c);
point(b);
plot([c(1) b(1)], [c(2) b(2)], 'k');
circle(a, r);
circle(P, (s-t));
axis equal
figure(gcf);


function point(pt)
hold on;
plot(pt(1), pt(2), 'ro');
hold off;



function circle(ctr, radius)
t = linspace(0, 2*pi, 100); 
x = ctr(1) + radius * cos(t);
y = ctr(2) + radius * sin(t); 
hold on;
plot(x, y); 
point(ctr);
hold off;

The blue circle is the one around point $A$; the red-orange circle is the computed on. The black line segment goes from $C$ to $B$.

Of course, you still have to work through the case where $t > s$ to find the coordinates of the center $P'$ and the radius of the second circle.

Geometric solution

Locus of points equidistant to a circle and a point (exterior to the circle) is a hyperbola.

These equidistant points are centers of circles through the given point, and touching the given circle.

In the present case, the given blue circle is denoted $gamma$ and is centered at $A.$ The hyperbola has foci $A$ and $B$ and passes through the middle point of the segment $BG,$ where $G$ is intersection of $AB$ and $gamma.$

Since the centers of circles touching $gamma$ must lie at $BC,$ they are at the intersection of $BC$ and the hyperbola.

The picture gives two possible configurations assuming that $BC$ has empty intersection with the circle:

1. the line cuts each branch of the hyperbola at a single point


2. the line cuts one branch of hyperbola at two points

(analytic solution can be added à la commande)

You haven’t mentioned this explicitly, but from the illustration is appears that you’re only interested in externally tangent circles. If the radius of the circle is $r$, then the points you are looking for satisfy $|PA|-|PB|=r$. Set $P = (1-t)B+tC$ and use the distance formula to get a somewhat messy-looking equation in $t$. You can find some useful suggestions for how to manipulate equations involving sums of radicals into a more manageable form in the answers to this question.