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.
Geometric solution
Locus of points equidistant to a circle and a point (outside 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:
(analytic solution can be added à la commande)
A strong hint, but not a complete solution
Let $$ \newcommand{\bu} {{\mathbf u}} \newcommand{\bv} {{\mathbf v}} \begin{align} s &= \|B - C\|\\ \bu &= \frac{B - C}{s} \end{align} $$ The point $P$ is then $C + t\bu$ 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 \begin{align} (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 & \text{algebra}\\ 2t \bu \cdot \bv - 2st -2rt &= e^2 -s^2 -2rs - r^2& \text{algebra}\\ t( -2 \bu \cdot \bv + 2s + 2r) &= (s+r)^2 - e^2& \text{algebra}\\ t &= \frac{(s+r)^2 - e^2}{-2 \bu \cdot \bv + 2s + 2r }& \text{algebra}\\ \end{align} ...so that gives you the point $P$ (you just compute $P + t\bu$). 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.
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.
