Right-angled triangle inside circle, chord as hypotenuse 
A 3-4-5 right triangle lies inside the circle $2x^2+2y^2=25$. The triangle is moved inside the circle in such a way that its hypotenuse always forms a chord of the circle. Find the locus of the vertex opposite to the hypotenuse.

Well I'm sort of confused on this question. I tried using polar coordinates and trig-bashing but its not giving suitable results. I am intuitively guessing that the locus will be a circle but I'm not sure.
 A: Hint. Note that the distance of the chord from the center of the circle is $5/2$, and the height of the triangle with respect to the chord is $12/5$. Moreover, this height divides  the chord into two segments of lengths $9/5$ and $16/5$. Hence the distance of the right-angle vertex from the center of the circle is the hypotenuse of the right triangle of sides $16/5-5/2=7/10$ and $5/2-12/5=1/10$, that is $\sqrt{7^2+1^2}/10=1/\sqrt{2}$.
Therefore the desired locus is a circle concentric with the original circle with radius $1/\sqrt{2}$.
A: 
The circle has radius $\frac5{\sqrt2}$. Suppose we position the triangle and circle such that


*

*the triangle hypotenuse is centred at the origin and parallel to the $x$-axis, forming a chord of the circle

*the circle centre and right-angled vertex are in the upper half-plane


Then the circle centre has coordinates $\left(0,\sqrt{\left(\frac5{\sqrt2}\right)^2-\left(\frac52\right)^2}\right)=\left(0,\frac52\right)$, while the right-angled vertex's coordinates are not much harder:

In other words, the right-angled vertex's coordinates are $\left(\frac7{10},\frac{12}5\right)$. The distance between the two vertices of interest, which remains the same no matter how the triangle is arranged in the circle, is thus
$$\sqrt{\left(\frac7{10}-0\right)^2+\left(\frac{12}5-\frac52\right)^2}=\frac{\sqrt2}2$$
and the locus of the right-angled vertex is thus $x^2+y^2=\frac12$.
