Using the coordinates of two of a triangle's vertices to calculate the coordinates of the third - trigonometry

Question

I know two coordinates of two vertices in a triangle (not aligned to an axis) and I'm attempting to calculate the coordinates of the third.
a
B ------- C
\ |
\ |
C' \ |
c \ | b
\ |
\ |
\|
A
I know the coordinates of A and B, the lengths of a and c, and that the angle C will always be a right angle. I believe there can only be two possible solutions for the coordinates of C; the one drawn above, and one with C reflected about the line c, approximately at C'. I'd like to calculate both positions.
EDIT:
The source of the triangle is as below.
I know the apex A, the centre of the circle B, the radius of the circle (a) and, from Pythag with (B - A), I know the length of c. I'm trying to find the points at which a line from the apex are at a tangent to each side of the circle, C and C'.
This appears to be an answer to my problem; can anyone elaborate on 'Given two sides of a right triangle, it's easy to find the length and direction of the third side.'.

Answer 1

I know the coordinates of A and B, and the lengths of a and c. From this, I believe there can only be two possible solutions for the coordinates of C
This is not true. There are an infinite number of choices for the position of C, as you don't know the length of b.
For example:
C
|
|
|
|
|
B
\
\
\
c \
\
\
\
A
If you connect C to A, you still maintain those known lengths....
In order for this to be true, you would also need to know one of the angles (such as that it's a right triangle), or the length of b.

Answer 2

It is easy: As the C angle is PI/2 then b=sqrt(c*c-a*a) so you know the lengths of a, b, c.
The coordinates of C and C' = the intersections of two circles:
center B radius a
center A radius b
Solved here for example //A=P1, B=P0, C=P3: https://math.stackexchange.com/questions/187107/calculate-coordinates-of-3rd-point-vertex-of-a-scalene-triangle-if-angles-and
The essential condition for this: if (a<c) otherwise it has no solution.

Answer 3

If you know it's going to be a right triangle, then you know the x and y values will be taken from the other two points.
Point coordsForCompletingTriangleTop(Point a, Point b) {
return new Point(a.x,b,y);
}
Point coordsForCompletingTriangleBottom(Point a, Point b) {
return new Point(b.x,a,y);
}
If cannot be guaranteed that it will be a right triangle, then you do need more information. The length of B, the length of C, or the angle of BCA would be required.

Answer 4

If you assume a and b are the opposite corners of a rectangle
a = (xa, ya)
b = (xb, yb)
then the top right rectangle point is c1 = (max(xa,xb), max(ya,yb))
and the bottom left rectangle point is c2 = (min(xa,xb), min(ya,yb))
Assuming that xa != xb and ya != yb
(xa, ya) A C1 (max(xa, xb), max(ya, yb))
o----------o
|\ |
| \ |
| \ |
| \ |
| \ |
| \ |
| \ |
| \ |
| \ |
o----------o
(min(xa, xb), min(ya, yb)) C2 B (xb, yb)
If your diagonal is going the other way (to test this see if xa > xb) you need to swap min for max on the x
(min(xa, xb), max(ya, yb)) C3 A'
o----------o
| /|
| / |
| / |
| / |
| / |
| / |
| / |
| / |
| / |
o----------o
B' C4 (max(xa, xb), min(ya, yb))
And if you're interested, the full set of solutions actually lies on the circle:
To compute this, suppose we have two points A = (xa, ya) and B = (xb, yb). Then the center point of this circle is c = (0.5 (xa + xb), 0.5 (ya + yb)) - just the midpoint of the A and B. The radius of the circle is r = sqrt( (xb - xa)^2 + (yb - ya)^2) / 2 - using pythagoras' theorem to get the length of the line and halving it. Then any point on the circle can be defined by p = c + (rcos(u), rsin (u)) for some angle u. There are 2 angles which give you the points p = A and p = B so these values of u are not good solutions. You can write out the equation and solve it for these 2 points to give you the values of u which you cannot use.