Given coordinates of hypotenuse, how can I calculate coordinates of other vertex? I have the Cartesian coordinates of the hypotenuse 'corners' in a right angle triangle. I also have the length of the sides of the triangle. What is the method of determining the coordinates of the third vertex where the opposite & adjacent sides meet.
Thanks, Kevin.
 A: You have two points $A=(a,b)$ and $B=(c,d)$ and want a point $P$
at given distances from $A$ and $B$, say $l$ and $m$. Then $|PA|^2=l^2$
and $|PB|^2=m^2$ that is
$$(x-a)^2+(y-b)^2=l^2\qquad\qquad(1)$$
and
$$(x-c)^2+(y-d)^2=m^2.\qquad\qquad(2)$$
Subtracting (2) from (1) gives a linear equation. Use this to eliminate
one variable from (1). This yields a quadratic equation in the other variable.
Solving this will give the two possible positions for $P$.
A: While I'd use the same algebra as Robin Chapman's solution, my first thought on this problem yields a third circle equation.  The circumcenter of a right triangle is at the midpoint of its hypotenuse.  Given the endpoints of the hypotenuse, $A=(a,b)$ and $B=(c,d)$, and letting $h$ be the length of the hypotenuse, the circumcircle has equation
$$\left(x-\frac{a+c}{2}\right)^2+\left(y-\frac{b+d}{2}\right)^2=\left(\frac{h}{2}\right)^2.$$
This may look a bit intimidating in symbols, but isn't really any different to work with for solving than the other two circles.  I don't know that this offers any advantages, though.
A: Let $AB$ the hypotenuse, let vector $\vec c=\overrightarrow{OB}-\overrightarrow{OA}$, its length $c$, the right angle at $C$, $DC=h$ the height of the triangle, $a$ and $b$ the given length of the legs, $q$ length of $AD$ as usual.  Define $J\colon R^2\rightarrow R^2$, $(v_1,v_2)\mapsto (-v_2,v_1)$ the rotation by 90 degree.
We know by Euclid that $q=b^2/c$ and elementarily that $ab=ch$, so $h=ab/c$. Then $\vec c/c$ is the unit vector of $c$, thus one solution is
$$\overrightarrow{OC}=\overrightarrow{OA}+\frac{b^2}{c}\frac{\vec c}{c}+\frac{ab}{c}J\Bigl(\frac{\vec c}{c}\Bigr)=\overrightarrow{OA}+\frac{b}{c^2}\bigl(b\vec c+ aJ(\vec c)\big).$$
Can you find the second solution?
Moral: Avoid coordinates!
Michael
