<s>[INST] Locus of vertex of triangle moving inside circle A right triangle with sides $3,4$ and $5$ 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. The locus of the vertex opposite to the hypotenuse is ____ ?
A) $2x^2+2y^2=1$
B) $x^2+y^2=1$
C) $x^2+y^2=2$
D) $2x^2+2y^2=5$
From the centre of circle distance to chord is $\displaystyle\frac{5}{2}$ units. From the vertex of triangle, distance to hypotenuse is $2.4$ units. However I am unable to sum up these two parts to obtain the final answer.
 [/INST] Some of the results are summarized into the following figure.

$θ = \angle DAX - \angle CAX = sin^{-1} \dfrac {2.5}{2.5√2} – sin^{-1} \dfrac {2.4}{4} = 8.130xxxxxx^0$
In ⊿CAD, $r^2$, the square of the required radius = … by cosine law … = 0.5
Thus, the required equation of the circle is $2x^2 + 2y^2 = 1$

Note:- The strange thing is that such a complicated problem is classified as an MC question. There must be some other easier method in finding the solution.
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