Locus of vertex of triangle moving inside circle

Question

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.

Answer 1

You can do a construction using ruler and compass, and you will notice that the distance between the center of the circle anf the right-angled corner of the triangle is less than one. Only one of your four possible answers agrees with this situation.

If you prefer to compute this distance, here is a useful illustration:

Since $AB$ with length $5$ is a diagonal in the square $AFBD$, the edges $AD$ and $BD$ are both of length $\frac52\sqrt2$, so $D$ is the center of the circle. You should be able to compute its coordinates, and from that the distance $CD$.

Answer 2

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 $2x^2 + 2y^2 = 1$

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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.