Let the circle be $\Omega$ and $R$ be the radius of $\Omega$. Let $\omega$ be the circle centered at $K$ with radius $R$. Let $M$ be the midpoint of $O$ and $K$, and let $\omega'$ be the circle centered at $M$ with radius $\frac R2$. I claim that the locus is $\omega'$. [asy][asy]size(200);pair O=(0,0),K=(2,0),P=(2+2sqrt(2),1); draw(circle(O,3)); draw(circle(K,3)^^circle(P,3)^^O--P,dotted); pair X[] = intersectionpoints(circle(O,3),circle(P,3)); draw(X[0]--X[1],dotted); dot((X[0]+X[1])/2,red); draw(circle(K/2,3/2),red); dot(O^^K^^K/2^^P);label("$P$",P,N); label("$O$",O,S);label("$K$",K,S);label("$M$",K/2,S);[/asy][/asy] Proof: It suffices to show that any point of the locus must be a midpoint of $O$ and some point on circle $\omega$. Let $P$ be the center of any arbitrary circle passing through $K$. Since the radius of this circle is $R$, this means $P$ lies on the circle centered at $K$ with radius $R$. Furthermore, the midpoint of the common chord of the two circles is also the midpoint of $OP$. Therefore, this midpoint lies on $\omega'$ This shows that the locus is indeed $\omega'$.

Let the first and second circles be $X_{1}$ and $X_{2}$ and let their common chord be $AB$. Now because of symmetry the reflection of point $K$ (let that be point $C$) by the line $AB$ lies on $X_{1}$. Now $XACB$ is a parallelogram and so the locus of the midpoint of $AB$ is the locus of the midpoint of $KC$. Take a homothety $h\left(K,\frac{1}{2}\right)$. Obviously, the locus is $h(X_{1})$ and we are done