The lines have equations $ y=2x+t,\frac{1}{2}x+t, $ where $ t $ is the ordonate of $ T. $ The points of intersection of these lines with that hyperbola are, after solving the two quadratic equations: $$ P_1=\left( \frac{-t+\sqrt{t^2+8}}{4} ,\frac{t+\sqrt{t^2+8}}{2} \right) ,P_2=\left( \frac{-t-\sqrt{t^2+8}}{4} ,\frac{t-\sqrt{t^2+8}}{2} \right) , $$$$ P_3=\left( -t-\sqrt{t^2+2} ,\frac{t-\sqrt{t^2+2}}{2} \right) ,P_4=\left( -t+\sqrt{t^2+2} ,\frac{t+\sqrt{t^2+2}}{2} \right) $$The equation of the circle passing through $ P_1,P_2,P_3 $ is, then, $$ 2\left( 4x+5t^3+5t+5t^2\sqrt{t^2+2} \right)^2 +\left( 8y-5t^3-8t-5t^2\sqrt{t^2+2} \right)^2 = $$$$ =250t^6+900t^4+885t^2+160+2t(5t^2+4)(25t^2+49)\sqrt{t^2+2} . $$$ P_4 $ should satisfy the above equation. Now, the locus of centers required should be the curve $ c:\mathbb{R}\longrightarrow\mathbb{R}^2 , $ defined as $$ c(t)=\left( -\frac{5t^3+5t+5t^2\sqrt{t^2+2}}{4} , \frac{5t^3+8t+5t^2\sqrt{t^2+2}}{8} \right) . $$It is possible that I might have erred some calculations, but the method is pretty straightforward.

We have lines $y=2x+t,y=\frac{1}{2}x+t$ Let $A(a,\frac{1}{a}),C(c,\frac{1}{c})$ are point of intersection of $y=2x+t$ with hyperbola and $B(b,\frac{1}{b}),D(d,\frac{1}{d})$ are point of intersection of $y=\frac{1}{2}x+t$ with hyperbola Then $ac=-\frac{1}{2},a+c=-\frac{t}{2}$ and $bd=-2,b+d=-2t$ Lemma: Lemma: Circumcenter of $A(a,\frac{1}{a}),B(b,\frac{1}{b}),C(c,\frac{1}{c})$ is $(\frac{a+b+c+\frac{1}{abc}}{2},\frac{abc+\frac{1}{a}+\frac{1}{b}+\frac{1}{c}}{2})$ Proof here $(a+b+c+\frac{1}{abc})-(a+b+d+\frac{1}{abd})= (c-d)(1-\frac{1}{abcd})=0$ and $(abc+\frac{1}{a}+\frac{1}{b}+\frac{1}{c})-(abd+\frac{1}{a}+\frac{1}{b}+\frac{1}{d})=(c-d)(ab-\frac{1}{cd})=\frac{(c-d)(abcd-1)}{cd}=0$ So circumcenters of $ABC$ and $ABD$ are same so $A,B,C,D$ lies on same circle $O_x=\frac{1}{2} (a+b+c+\frac{1}{abc})=\frac{1}{2} (a+b+c+d)=\frac{1}{2} (-\frac{t}{2}-2t)=-\frac{5t}{4}$ $O_y=\frac{1}{2} (abc+\frac{1}{a}+\frac{1}{b}+\frac{1}{c})=\frac{1}{2} (abc+bcd+cda+dac)=\frac{1}{2} ( ac(b+d)+bd(a+c))=t$ So circumcenter has coordinates $(t,-\frac{5t}{4})$ and so locus of centers, when point $T$ runs through the entire $y$-axis is line $ y=-\frac{5}{4}x$