First suppose $AA_1$, $BB_1$, and $CC_1$ do not concur. [asy][asy] size(10cm); pair A = dir(100); pair B = dir(220); pair C = dir(-40); pair A1 = B+0.4*(C-B); pair B1 = C+0.6*(A-C); pair C1 = A+0.3*(B-A); pair D = extension(B, B1, C, C1); pair E = extension(C, C1, A, A1); pair F = extension(A, A1, B, B1); pair A2 = 2*foot(origin, A, A1)-A; pair B2 = 2*foot(origin, B, B1)-B; pair C2 = 2*foot(origin, C, C1)-C; pair Ax = 2*foot(origin, A2, D)-A2; pair Bx = 2*foot(origin, B2, E)-B2; pair Cx = 2*foot(origin, C2, F)-C2; draw(unitcircle); draw(A--B--C--cycle); draw(A--A2, dashed); draw(B--B2, dashed); draw(C--C2, dashed); draw(A2--Ax, dotted); draw(B2--Bx, dotted); draw(C2--Cx, dotted); draw(Ax--Bx--Cx--cycle, red); string[] names = {"$A$", "$B$", "$C$", "$A_1$", "$B_1$", "$C_1$", "$D$", "$E$", "$F$", "$A_2$", "$B_2$", "$C_2$", "$A'$", "$B'$", "$C'$"}; pair[] pts = {A, B, C, A1, B1, C1, D, E, F, A2, B2, C2, Ax, Bx, Cx}; pair[] labels = {A, B, C, dir(225), dir(0), dir(180), dir(0), E, F, A2, B2, C2, Ax, Bx, Cx}; for(int i=0; i<names.length; ++i){ dot(names[i], pts[i], dir(labels[i])); } [/asy][/asy] Let $D=BB_1\cap CC_1$, $E=CC_1\cap AA_1$, and $F=AA_1\cap BB_1$. Let $AA_1$, $BB_1$, and $CC_1$ meet $k$ again at $A_2$, $B_2$, and $C_2$. Let $A_2D$, $B_2E$, and $C_2F$ meet $k$ again at $A'$, $B'$, and $C'$. Note that $A'\neq A$. By Pascal's theorem on $B'C'C_2CBB_2$, $A_1$ lies on $B'C'$. Analogously $B_1$ and $C_1$ lie on $C'A'$ and $A'B'$ respectively, so we cannot restore the triangle because both $ABC$ and $A'B'C'$ are valid. In the other direction, suppose there were two possible triangles $ABC$ and $A'B'C'$. Let $T$ be the common concurrency point of $AA_1$, $BB_1$, and $CC_1$. Let $AT$ and $BT$ meet $k$ again at $A_2$ and $B_2$. By Pascal's theorem on $A'A_2ABB_2B'$, points $C_1$, $AT\cap B'B_2$, and $BT\cap A'A_2$ are collinear. (unfinished) Here is an explicit construction in the case where $AA_1$, $BB_1$, and $CC_1$ are concurrent (but this leaves open the possibility that there is a valid $A'B'C'$ with $A'A_1$, $B'B_1$, $C'C_1$ not concurrent). Let $X=B_1C_1\cap BC$. Then $(X, A_1; B, C)=-1$ so $X$ lies on the polar of $A_1$. This means we can reconstruct $X$ from $A_1$, $B_1$, and $C_1$ alone as the intersection of $B_1C_1$ with the polar of $A_1$. Then we can construct line $A_1X$, which is line $BC$. Analogously we can reconstruct lines $CA$ and $AB$, so triangle $ABC$ can be restored.

Solved (a while ago) with SwimWithDolphin. This solution doesn't leave open the possibility mentioned above Projective transform (uniquely, which allows us to apply an inverse transform and obtain the original triangle) $C_1$ to the center of $k$, while fixing $k$, so no matter what $\overline{AB}$ is a diameter. Let $\Gamma=(A_1B_1)$. If some working $\triangle ABC$ exists, then $\angle A_1CB_1=90^\circ$, so it must lie on $\Gamma$ as well. On the other hand, if $\Gamma$ intersects $k$ twice, then we can pick either of the intersection points and construct a valid $\triangle ABC$. Therefore a unique restoration exists if and only if $\Gamma$ is tangent to $k$. For homothety reasons, $\Gamma$ is tangent to $k$ if and only if in this unique restoration, we have $\overline{A_1B_1} \parallel \overline{AB}$, which by Ceva/similar triangles occurs if and only if $AA_1,BB_1,CC_1$ are concurrent. $\blacksquare$