Find locus of point vertex of right angle triangle whose hypotenuse is chord of circle. Let $AB$ be a variable chord of length 5 to the circle $$x^2+y^2=\frac{25}{2}$$. A triangle $ABC$ is constructed such that $BC=4$ and $CA=3$. If the locus of C is $x^2+y^2=a$ find all possible values of $a$
I tried using the parametric form to solve this question but it resulted in a very cumbersome task with no avail because of so much of variables and their eliminations. Can anyone explain any easier method to solve it. 

Consider the particular position of $A\left(-\frac52,\frac52\right);\;B\left(\frac52,\frac52\right)$
The line $AC_2$ has equation $y-\frac{5}{2}=\frac{4}{3} \left(x+\frac{5}{2}\right)$
And the circle with center $A$ and radius $3$ has equation $\left(x+\frac{5}{2}\right)^2+\left(y-\frac{5}{2}\right)^2=9$
The intersection point is $C_2(-0.7,\;4.9)$
Similarly for $AC_1$. The equation is $y-\frac{5}{2}=-\frac{4}{3} \left(x+\frac{5}{2}\right)$
Intersecting with the same circle center $A$ and radius $3$ we get
$C_1(-0.7,0.1)$
The radii of the two loci are 
$OC_1=\sqrt{0.7^2+0.1^2}=\frac{1}{\sqrt{2}}$
so the first value for $a_1=\frac12$
And $OC_2=\sqrt{0.7^2+4.9^2}=\frac{7}{\sqrt{2}}$
and we get $a_2=\frac{49}{2}$
Hope this helps
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