Find right triangle inscribed in circle from two points on separate legs I am getting started on problem solving, and need help with the following exercise of "Solving mathematical problems, a personal perspective". The full question is as follows:
"We are given a circle and two points A and B inside the circle. If possible, construct a right-angled triangle inscribed in the circle such that one leg of the right-angled triangle contains A and the other leg contains B; see figure below. (Hint: solve for the right-angled vertex.)"

I'm not familiar with these kinds of problems, and quite frankly don't know where to start. If you can provide an answer, could you please start it off with a hint, as I don't really want to be spoon-fed all the way. Thanks!
 A: If $C$ is the right angle vertex of a triangle that satisfies the problem statement, note that $\triangle ABC$ is also a right triangle with right angle at $C.$
Find the set of all points that can be the right-angled vertex of a right triangle whose other vertices are $A$ and $B.$
The point $C$ cannot be anywhere except in that set of points.
Of course, $C$ must also be on the given circle.
You now have two facts about $C$ which can help you to find that point.
(Note that there might be more than one choice or there might be none, depending on how the construction goes.)
A: Suppose the circle is of radius $R$ centered at the origin, and that point $A$ is $a$ units away from the center, and makes an angle of $\theta_1$ with the positive $x$ axis.  Similarly, point $B$ is $b$ units from the center, and makes an angle of $\theta_2$ from the positive $x$ axis.
$A = a (\cos \theta_1, \sin \theta_1 )$
$B = b (\cos \theta_2, \sin \theta_2)$
The vertex of the right angle is $V = R (\cos \phi, \sin \phi ) $
Condition is:
$(V - A ) \cdot (V - B ) = 0 $
Expanding,
$ V \cdot V - (A + B) \cdot V + A \cdot B = 0 $
This becomes,
$ R^2 - (A + B) \cdot V + A \cdot B = 0 $
Let $C = A + B$, then the last equation becomes,
$ R C_x \cos \phi + R C_y \sin \phi = R^2 + A \cdot B $
which has two solutions in $\phi$ as long as
$ \left| \dfrac{R^2 + A \cdot B }{R \sqrt{ (A + B) \cdot (A + B ) } } \right| \lt 1 $
Finally we have to find the other two vertices of the right triangle, so let point $A$ lie on $VU$, then
$U = V + t VA$
where we know that U \cdot U = R^2, therefore,
$ (V + t VA) \cdot (V + t VA) = R^2 $
But $V \cdot V = R^2$, and $t $ cannot be zero (this corresponds to $V$ itself), hence,
$ t = - 2 \dfrac{ V \cdot VA }{ VA \cdot VA } $
Finally the point $W$, such that $B$ lies on $VW$ is given by
$W = - U $
