The locus of the middle points of the chords of the circle $x^2+y^2=4a^2$ which subtend a right angle at the centre of the circle is 

Let $M(h, k)$  be the middle point of the chord $AB$ of the given circle

$x^2+y^2=4a^2$ , with centre at $O(0,0)$ 

and radius equal to $2a.$

  Then $OM$is perpendicular to AB

 Since $AOB$ is a right angled triangle 

$\begin{array}{l}4(AM{)}^2=(AB{)}^2=(OA{)}^2+(OB{)}^2\\ =(2a{)}^2+(2a{)}^2=8a^2\\ AM=\sqrt{2}a\end{array}$

Also   $(OA{)}^2=(OM{)}^2+(AM{)}^2$

        $\Rightarrow (2a{)}^2=h^2+k^2+2a^2$

       $\Rightarrow h^2+k^2=2a^2$

       $\Rightarrow$     Locus of $(h, k)$  is  $x^2+y^2=2a^2$