bah can someone post a synthetic solution please WLOG, diagram as shown. Let \(A'\) and \(B'\) lie on \(\Gamma\) such that \(\overline{AA'}\parallel\overline{U_1V_1}\) and \(\overline{BB'}\parallel\overline{U_2V_2}\). I claim the fixed point is \(K=\overline{AB'}\cap\overline{BA'}\). [asy][asy] size(6cm); defaultpen(fontsize(10pt)); pair A,B,K,X,C,D,L,Ap,Bp,U1,V1,U2,V2; A=dir(170); B=dir(10); K=0.3*dir(210); X=dir(80); D=(B+2X)/3; C=(3X+4A)/7; L=extension(K,D,X,A); Ap=2*foot(origin,B,K)-B; Bp=2*foot(origin,A,K)-A; U1=intersectionpoint(unitcircle,C--(C+100*(Ap-A))); U2=intersectionpoint(unitcircle,D--(D+100*(Bp-B))); V1=2*foot(origin,C,U1)-U1; V2=2*foot(origin,D,U2)-U2; draw(U1--V1,gray+Dotted); draw(U2--V2,gray+Dotted); draw(Ap--K--Bp,gray+dashed); draw(A--Ap,gray); draw(B--Bp,gray); draw(circumcircle(X,C,D),gray); draw(X--L--K,gray); draw(unitcircle); draw(A--X--B); draw(A--K--B,dashed); dot("\(A'\)",Ap,Ap,gray); dot("\(B'\)",Bp,Bp,gray); dot("\(U_1\)",U1,U1,gray); dot("\(V_1\)",V1,V1,gray); dot("\(U_2\)",U2,U2,gray); dot("\(V_2\)",V2,N,gray); dot("\(A\)",A,A); dot("\(B\)",B,B); dot("\(K\)",K,S); dot("\(L\)",L,NE); dot("\(X\)",X,N); dot("\(C\)",C,dir(150)); dot("\(D\)",D,dir(190)); [/asy][/asy] We will show that \(\operatorname{Pow}(K,(XCD))\) depends only on the locations of \(A\), \(B\), \(K\), and the distances \(\alpha\) and \(\beta\) from \(C\) and \(D\) to \(\overline{AA'}\) and \(\overline{BB'}\), respectively. To this end, define the function \(f:\mathbb R^2\to\mathbb R\) by \[f(\bullet)=\operatorname{Pow}(\bullet,(XCD))-\operatorname{Pow}(\bullet,\Gamma).\]It is known that \(f\) is linear. First, note that \begin{align*} \alpha&=AC\sin\angle CAA'=AC\sin\angle XBK\\ \beta&=BD\sin\angle DBB'=BD\sin\angle XAK. \end{align*} Denote \(L=\overline{AX}\cap\overline{DK}\). It is easy to compute that \begin{align*} f(L)&=LX\cdot LC-LX\cdot LA=-LX\cdot CA\\ f(D)&=-(-DB\cdot XD)=DB\cdot XD. \end{align*}Moreover, observe from the Law of Sines that \begin{align*} \frac{KL}{LD}&=\frac{AK\cdot\sin\angle XAK}{LD\cdot\sin\angle XLD}=\frac{AK\cdot\sin\angle XAK}{XD\cdot\sin\angle AXB}\\[1.5ex] \frac{KD}{LD}&=\frac{BK\cdot\sin\angle XBK}{LD\cdot\sin\angle LDX}=\frac{BK\cdot\sin\angle XBK}{LX\cdot\sin\angle AXB}. \end{align*}Since \(f\) is linear, we know \begin{align*} f(K)&=\frac{KL}{LD}\cdot f(D)-\frac{KD}{LD}\cdot f(L)\\[1.5ex] &=\frac{AK\cdot\sin\angle XAK\cdot DB+BK\cdot\sin\angle XBK\cdot CA}{\sin\angle AXB}\\[1.5ex] &=\frac{AK\cdot\beta+BK\cdot\alpha}{\sin\angle AXB}, \end{align*}which is dependent only on \(A\), \(B\), \(K\), \(\alpha\), \(\beta\). Since \(\operatorname{Pow}(K,\Gamma)\) is fixed, \(\operatorname{Pow}(K,(XCD))\) is also dependent only on \(A\), \(B\), \(K\), \(\alpha\), \(\beta\), so we are done.

TheUltimate123 wrote: WLOG, diagram as shown. Let \(A'\) and \(B'\) lie on \(\Gamma\) such that \(\overline{AA'}\parallel\overline{U_1V_1}\) and \(\overline{BB'}\parallel\overline{U_2V_2}\). I claim the fixed point is \(K=\overline{AB'}\cap\overline{BA'}\). Hmm.... I suppose that after guessing the location of $K$ and generalizing problem ,our work is easy with Animation method...($X$ moves on circle :2 ,and $C,D$ on lines :1+1 so we'll need just 5 points on circle)

Amazing... this problem is a geo problem from a relatively high-profile contest, and it has been on AoPS for over 2 hours with no synthetic solutions posted! It must be very hard synthetically

Here is a simple and very short synthetic solution I found while testsolving: Let $B'$ be on $\Gamma$ with $BB'||U_2V_2$ and $A'$ be on $\Gamma$ with $AA'||U_1V_1$. We claim the point $K$ is $AB' \cap BA'$. First, let $AB' \cap U_2V_2 = B_1$ and define $A_1$ similarly. Then $\angle AXB = 180^{\circ} - \angle AB'B = 180^{\circ} - \angle AB_1D$, so $XADB_1$ is cyclic. Similarly, $BXA_1C$ is cyclic. For a circle $\omega$ let $p_Q(\omega)$ be the power of $Q$ with respect to circle $\omega$. From our previous observations, it follows that $p_K((AXD))=KA\cdot KB_1$, and that $p_K ((BXC))=KB\cdot KA_1$, which are both fixed. We also obviously have that $p_K((AXB))$ is fixed because $(AXB)$ does not change. It now follows that $p_K ((CXD))$ is fixed, because for all points $Q$ we have $p_Q ((AXB)) + p_Q ((CXD)) = p_Q ((AXD)) + p_Q ((BXC)) \blacksquare$

We drop the condition that $X$ must lie between $V_1$ and $V_2$. It suffices to show that as $X$ varies, the pairwise radical axes of the circles $(XCD)$ all pass through some fixed point $K$.

Let $A',B'\in\Gamma$ such that $AA'\parallel U_1V_1$ and $BB'\parallel U_2V_2$. Let $I\in U_1V_1$ such that $DI\parallel A'B$ and $J\in U_2V_2$ such that $CJ\parallel AB'$. We have $I,J\in(XCD)$ by Reim's theorem. All distances will be signed in the most natural way. Let $t=V_1C$ (negative if $C$ on the other side of $V_1$), and let $s=V_2D$. Note that $I$ varies linearly as $D$ varies, so $V_1I = ps+q$ for some fixed $p,q$, where $p\ne 0$. Thus, \[\mathrm{pow}_{(XCD)}(V_1) = V_1C\cdot V_1I = t(ps+q) = \alpha_1 ts + \beta_1 t+\gamma_1 s+\delta_1\]for some fixed $\alpha_1,\beta_1,\gamma_1,\delta_1$. Similarly we can show that \[\mathrm{pow}_{(XCD)}(V_2) = \alpha_2 ts + \beta_2 t+\gamma_2 s+\delta_2\]and \[\mathrm{pow}_{(XCD)}(U_1) = \alpha_3 ts + \beta_3 t+\gamma_3 s+\delta_3,\]where $\alpha_i,\ldots,\delta_i$ are all fixed real numbers. We will now work in barycentric coordinates with respect to $\triangle V_1V_2U_1$ (and implicitly thus the projective plane). Let \[f(P) = \mathrm{pow}_{(XCD)}(P) - \mathrm{pow}_\Gamma(P).\]It is well known that $f$ is linear. Let $v_i=(\alpha_i,\beta_i,\gamma_i)\in\mathbb{R}^3$. We now have two cases. Case 1: Suppose the vectors $v_1,v_2,v_3$ are linearly dependent in $\mathbb{R}^3$. Then, there exist $a,b,c$ not all $0$ such that \[av_1+bv_2+cv_3=0.\]Pick $K=(a:b:c)$ (it is ok if $a+b+c=0$). Then, \[f(K) = a\delta_1+b\delta_2+c\delta_3,\]a constant, so $K$ passes through all the common radical axes. Case 2: Suppose the vectors $v_1,v_2,v_3$ are linearly independent in $\mathbb{R}^3$. We will need an important lemma. Lemma: There exist fixed real numbers $p,q,r$, not all $0$, such that $pts+qt+rs$ is a constant as $X$ varies. Proof: Note that the map $C\mapsto X\mapsto D$ is projective, so there exist constants $b_1,b_2,b_3,b_4$ with $b_1b_4\ne b_2b_3$ such that \[s = \frac{b_1t+b_2}{b_3t+b_4}.\]It is easy to check now that $p=-b_3$, $q=b_1$, $r=-b_4$ work. $\blacksquare$ Now, since $v_1,v_2,v_3$ are linearly independent, they must span $\mathbb{R}^3$, so select $a,b,c$ such that \[(p,q,r) = av_1+bv_2+cv_3.\]Note that $a,b,c$ are not all $0$ since $p,q,r$ are not all $0$. Again, let $K=(a:b:c)$, and now \[f(K) = pts+qt+rs + a\delta_1+b\delta_2+c\delta_3,\]again a constant, so we're done. Thus, we've found a point $K$ in the projective plane which passes through all the pairwise radical axes of $(XCD)$ as $X$ varies. All we have to show now is that $K$ is not an infinity point. If $K$ was an infinity point, then all the centers of the $(XCD)$s would be collinear, and in particular if $(XCD)$ was ever a line, it would only have one possible direction. However, it is easy to check that as $X\to A'$, $(XCD)$ degenerates to the line $A'B$, and as $X\to B'$, it degenerates to the line $AB'$. The lines $A'B$ and $AB'$ are not parallel, so we have the desired contradiction.

How did you find K? What did you do to find it? It seems so hard to find it.

SerdarBozdag wrote: How did you find K? What did you do to find it? It seems so hard to find it. Choose a few specific values of $X$, in the notation of tasty's post, $X \to A'$ and $X \to B'$ should suffice.

Here was the proposer's solution (completely synthetic). For brevity, we let $\ell_i$ denote line $U_iV_i$ for $i=1,2$. We first give an explicit description of the fixed point $K$. Let $E$ and $F$ be points on $\Gamma$ such that $\overline{AE} \parallel \ell_1$ and $\overline{BF} \parallel \ell_2$. The problem conditions imply that $E$ lies between $U_1$ and $A$ while $F$ lies between $U_2$ and $B$. Then we let \[ K = \overline{AF} \cap \overline{BE}. \]This point exists because $AEFB$ are the vertices of a convex quadrilateral. Let $Y$ be the second intersection of $(XCD)$ with $\Gamma$. Let $S = \overline{EY} \cap \ell_1$ and $T = \overline{FY} \cap \ell_2$. Claim: Points $S$ and $T$ lies on $(XCD)$ as well. Proof. By Reim's theorem: $\measuredangle CSY = \measuredangle AEY = \measuredangle AXY = \measuredangle CXY$, etc. $\blacksquare$ Now let $X'$ be any other choice of $X$, and define $C'$ and $D'$ in the obvious way. We are going to show that $K$ lies on the radical axis of $(XCD)$ and $(X'C'D')$. [asy][asy] size(8cm); pair A = dir(190); pair B = -conj(A); pair U_1 = dir(125); pair V_1 = dir(220); pair U_2 = dir(80); pair V_2 = dir(310); filldraw(unitcircle, invisible, blue); draw(A--B, blue); draw(U_1--V_1, deepgreen); draw(U_2--V_2, deepgreen); pair E = V_1*U_1/A; pair F = V_2*U_2/B; pair X = dir(265); pair C = extension(A, X, U_1, V_1); pair D = extension(B, X, U_2, V_2); draw(A--X--B, lightred); draw(A--E, deepcyan); draw(B--F, deepcyan); pair K = extension(A, F, B, E); draw(A--F, blue); draw(B--E, blue); pair Y = -X+2*foot(X, origin, circumcenter(X, C, D)); draw(circumcircle(X, C, D), red); pair S = extension(Y, E, U_1, V_1); pair T = extension(Y, F, U_2, V_2); draw(E--Y--F, brown); pair Xp = dir(300); pair Cp = extension(Xp, A, U_1, V_1); pair Dp = extension(Xp, B, U_2, V_2); draw(A--Xp--B, orange); pair L = extension(S, Y, Cp, Xp); dot("$A$", A, dir(A)); dot("$B$", B, dir(B)); dot("$U_1$", U_1, dir(U_1)); dot("$V_1$", V_1, dir(V_1)); dot("$U_2$", U_2, dir(U_2)); dot("$V_2$", V_2, dir(V_2)); dot("$E$", E, dir(E)); dot("$F$", F, dir(F)); dot("$X$", X, dir(X)); dot("$C$", C, dir(C)); dot("$D$", D, dir(D)); dot("$K$", K, dir(270)); dot("$Y$", Y, dir(Y)); dot("$S$", S, dir(S)); dot("$T$", T, dir(T)); dot("$X'$", Xp, dir(Xp)); dot("$C'$", Cp, dir(170)); dot("$D'$", Dp, dir(10)); dot("$L$", L, dir(L)); /* TSQ Source: A = dir 190 B = -conj(A) U_1 = dir 125 V_1 = dir 220 U_2 = dir 80 V_2 = dir 310 unitcircle 0.1 lightcyan / blue A--B blue U_1--V_1 deepgreen U_2--V_2 deepgreen E = V_1*U_1/A F = V_2*U_2/B X = dir 265 C = extension A X U_1 V_1 D = extension B X U_2 V_2 A--X--B lightred A--E deepcyan B--F deepcyan K = extension A F B E R270 A--F blue B--E blue Y = -X+2*foot X origin circumcenter X C D circumcircle X C D red S = extension Y E U_1 V_1 T = extension Y F U_2 V_2 E--Y--F brown X' = dir 300 C' = extension Xp A U_1 V_1 R170 D' = extension Xp B U_2 V_2 R10 A--Xp--B orange L = extension S Y Cp Xp */ [/asy][/asy] The main idea is as follows: Claim: The point $L = \overline{EY} \cap \overline{AX'}$ lies on the radical axis. By symmetry, so does the point $M = \overline{FY} \cap \overline{BX'}$ (not pictured). Proof. Again by Reim's theorem, $SC'YX'$ is cyclic. Hence we have \[ \operatorname{pow}(L, X'C'D') = LC' \cdot LX' = LS \cdot LY = \operatorname{pow}(L, XCD). \]$\blacksquare$ To conclude, note that by Pascal theorem on \[ EYFAX'B \]it follows $K$, $L$, $M$ are collinear, as needed.

Another way. Let $X_1,X_2,X_3,X_4$ be the four distinct points on arc $V_1V_2$ and $X_iA\cap U_1V_1=C_i, X_iB\cap U_2V_2=D_i$. Also, define $\omega_i$ be the circumcircle of triangle $X_iC_iD_i$. It suffices to show that $\omega_1,\omega_2,\omega_3,\omega_4$ have radical center. $\cdots (\#)$ Here, let $(i,j)$ be any pair of positive integers less than or equal to $4$. Let $\omega_i \cap U_1V_1=E_i(\neq C_i),\omega_i\cap U_2V_2=F_i(\neq D_i)$ and we call the lines $C_iF_i,D_iE_i$ by $l_i,g_i$ respectively. Denote $l_i\cap g_j=Y_{i,j}$. From simple angle chasing, $C_iE_iD_jF_j$ is cyclic so $Y_{i,j}$ is radical center of $\omega_i,\omega_j,(C_iE_iD_jF_j)$. Thus $Y_{i,j}Y_{j,i}$ is radical axis of $\omega_i,\omega_j$ and then $(\#)$ is equivalent to $Y_{3,1}Y_{1,3},Y_{2,3}Y_{3,2},Y_{2.4}Y_{4,2}$ are concurrent. Now since $\measuredangle C_iF_iD_i=\measuredangle AX_iB=\measuredangle AX_jB=\measuredangle C_jF_jD_j,$ it follows $l_i\parallel l_j$. Similarly, $g_i\parallel g_j$ hence $Y_{i,j}Y_{i+1,j}Y_{i+1,j+1},Y_{i,j+1}$ is parallelogram, especially $Y_{i,j}Y_{i,j+1}=Y_{i+1,j}Y_{i+1,j+1}$. Moreover, from $ABX_1X_2X_3X_4$ is cyclic, $$(l_1,l_2;l_3,l_4)=(C_1,C_2;C_3,C_4)=(D_1,D_2;D_3,D_4)=(g_1,g_2;g_3.g_4)$$so $$\frac{Y_{2,3}Y_{2,4}}{Y_{2,2}Y_{2,4}}\cdot\frac{Y_{3,3}Y_{3,1}}{Y_{3,2}Y_{3,1}}=\frac{Y_{3,2}Y_{4,2}}{Y_{2,2}Y_{4,2}}\cdot\frac{Y_{3,3}Y_{1,3 }}{Y_{2,3}Y_{1,3}},$$implying that $Y_{1,3}Y_{3,1},Y_{2,3}Y_{3,2},Y_{2,4}Y_{4,2}$ are concurrent, as desired.

Let $A', B'$ be points on $\Omega$ such that $AA' \parallel U_1V_1, BB' \parallel U_2V_2$. Then $T = AB' \cap A'B$ is the desired point. The proof is trivial, as by linearity of power of a point it suffices to show $\text{Pow}_{(XCD)}(T)-\text{Pow}_{\Omega}(T)$ is constant. If $P_A, P_B$ are the point corresponding to the components of $T$ in a vector space with basis vectors $XA$ and $XB$, then it suffices to show $AC \cdot AX \cdot \frac{XP_A}{XA}$ is constant over all choices of $A$ since by symmetry the same will be true for $B$ and thus their sum, the desired power difference, would be constant. This value is indeed constant where $\theta = \angle A'AX, \alpha = \angle ABX, A'' = A'B \cap AX, B'' = AB' \cap BX$, and $R$ is the radius of $\Omega$ since $AC \cdot AX \cdot \frac{XP_A}{XA} = \frac{d(AA', U_1V_1)}{\sin \theta} \cdot 2R \sin \alpha \cdot \frac{B''T}{B''A} = 2R \cdot d(AA', U_1V_1) \cdot \frac{\sin \alpha}{\sin \theta} \cdot \frac{BT}{BA} \cdot \frac{\sin \theta}{\sin \alpha} = 2R \cdot d(AA', U_1V_1) \cdot \frac{BT}{BA}$ which does not depend on $X$, so we are done.

Let $A'$ denote the point on $\Gamma$ such that $\overline{AA'} \parallel \overline{U_1V_1}$, and similarly define $B'$. We specialize the choice $K=\overline{AB'} \cap \overline{A'B}$. I claim that this satisfies the problem statement. Lemma: For any $\bullet$, we have \[ \text{Pow}(\bullet, (XBA))-\text{Pow}(\bullet, (XDA))=\text{Pow}(\bullet, (XBC))-\text{Pow}(\bullet, (XDC)). \]Proof. Let $f(\bullet)$ denote LHS and $g(\bullet)$ denote the RHS. Note that $f$ and $g$ are linear functions. It suffices to show that $f(\bullet)=g(\bullet)$ holds for three distinct non-collinear positions of $\bullet$, since every point in the plane can be described a linear combination of those points. This task is easy: choose $\bullet = X, C, D$. It is easy to verify that each of these points all work, by the means of radical axis. Claim: The powers of $K$ with respect to $(XBC)$ and $(XAD)$ are each independent of $X$. Proof. Let $P=\overline{A'B} \cap \overline{U_1V_1}$ and $Q=\overline{AB'} \cap \overline{U_2V_2}$. I contend that $P$ lies on $(XBC)$ and $Q$ lies on $(XAD)$. To show this, we angle chase and find \[ \angle BPC = \angle BA'A = 180^{\circ} - \angle BXA = 180^{\circ} - \angle BXC, \]which implies the concyclicity for $P$; the concyclicity for $Q$ is analogous. Thus, the power of $K$ with respect to $(XBC)$ is $-KB \cdot KP$, which is clearly independent of $X$, and similarly with $(XAD)$. Implementing the lemma, we note that by the claim, it suffices to show that the power of $K$ with respect to $(XAB)$ is independent of $X$, but $(XAB)$ is precisely $(ABC)$, so we are done.

Let $A_1$ and $B_2$ be the points on $\Gamma$ so that $AA_1 \parallel U_1V_1$ and $BB_2 \parallel U_2V_2$. We will show that the desired fixed point is $AB_2 \cap A_1B = Y$. First we show that $Pow_{(XCD)}(Y) - Pow_{(XAB)}(Y) = Pow_{(XBC)}(Y) - Pow_{(XAD)}(Y)$ which actually holds for any point, not just $Y$. By Linearity of Power of a Point it suffices to show that it holds true for some $3$ points. Clearly this holds true if the point is $X$ since all of the powers equal $0$. This is also true if the point is $C$ since $Pow_{(XCD)}(C) - Pow_{(XAB)}(C) = Pow_{(XBC)}(C) - Pow_{(XAD)}(C) = -CX \cdot CA$ and similarly for $D$ so we are done. Since $(XAB)$ is fixed it now suffices to show that $Pow_{(XBC)}(Y) - Pow_{(XAD)}(Y)$ is fixed. Let $Z = U_1V_1 \cap A_1B$. Then $\measuredangle BXC = \measuredangle BA_1A = \measuredangle BZC$ so $BXCZ$ is cyclic. Then $Pow_{(XBC)}(Y) = YZ \cdot YB$ which is fixed as $Z$ is fixed. So then similarly $Pow_{(XAD)}(Y)$ is fixed so we are done.

We will work in the complex projective plane $\mathbb{CP}^2$, and let $I=(1:i:0)$ and $J=(1:-i:0)$ be the circle points, which we can verify that every circle goes through, as any circle has equation $(x-az)^2+(y-bz)^2=(cz)^2$ and $(1-a(0))^2+(i-b(0))^2=(c(0))^2=0$ and $(1-a(0))^2+(-i-b(0))^2=(c(0))^2=0$. We will let a general conic with equation $Ax^2+By^2+Cz^2+Dyz+Ezx+Fxy$ be represented $(A:B:C:D:E:F)$ in $\mathbb{CP}^5$. In this way, given a conic varying in terms of a single variable $t$, we can let the degree of the conic be the minimum of the maximum degree of polynomials $A(t),B(t),C(t),D(t),E(t),F(t)$ where the conic can be represented as $(A(t):B(t):C(t):D(t):E(t):F(t))$ in $\mathbb{CP}^5$ over all possible $A(t),B(t),C(t),D(t),E(t),F(t)$, when such polynomials exist. Lemma $1$. Let $P_1,P_2,P_3,P_4,$ and $P_5$ be points varying with degrees $d_1,d_2,d_3,d_4,$ and $d_5$, respectively, with respect to a single variable $t$. Let $x$ be the number of values of $t$ for which at least $2$ of $P_1,P_2,P_3,P_4,$ and $P_5$ are the same. Then, the degree of the conic through $P_1,P_2,P_3,P_4,$ and $P_5$ is at most $2\left(d_1+d_2+d_3+d_4+d_5\right)-x$. Proof. Let the conic be $Ax^2+By^2+Cz^2+Dyz+Ezx+Fxy=0$, where $A,B,C,D,E,$ and $F$ are functions of $t$. We see that if $P_k=\left(x_k:y_k:z_k\right)$, where $x_k,y_k.$ and $z_k$ are functions of $t$ for $k=1,2,3,4,5$, then we get that if $$v_k=\begin{pmatrix}x_k^2\\y_k^2\\z_k^2\\y_kz_k\\z_kx_k\\x_ky_k\end{pmatrix}$$for $k=1,2,3,4,5$ then if $$v=\begin{pmatrix}A\\B\\C\\D\\E\\F\end{pmatrix}$$we have that $$v\bullet{}v_k=0$$for $k=1,2,3,4,5$, where $\bullet$ represents the dot product. as $P_k$ is on the conic for $k=1,2,3,4,5$. Therefore, we see that we can have $v$ equal to $$\begin{pmatrix}\left|M_1\right|\\-\left|M_2\right|\\\left|M_3\right|\\-\left|M_4\right|\\\left|M_5\right|\\-\left|M_6\right|\end{pmatrix}$$where if $$M=\begin{pmatrix}\text{---}v_1\text{---}\\\text{---}v_2\text{---}\\\text{---}v_3\text{---}\\\text{---}v_3\text{---}\\\text{---}v_4\text{---}\\\text{---}v_5\text{---}\end{pmatrix}$$then $M_k$ is the matrix resulting from deleting the $k$th column of $M$ for $k=1,2,3,4,5,6$ as then if $v_{k,j}$ is the $j$th coordinate of $v_k$ for $k=1,2,3,4,5$ and $j=1,2,3,4,5,6$ then \begin{align*} v\bullet{}v_k&=\sum^6_{j=1}(-1)^{j+1}v_{k,j}\left|M_k\right|\\ &=\begin{pmatrix}\text{---}v_k\text{---}\\\text{---}v_1\text{---}\\\text{---}v_2\text{---}\\\text{---}v_3\text{---}\\\text{---}v_4\text{---}\\\text{---}v_5\text{---}\end{pmatrix}\\ &=0 \end{align*}since $v_1,v_2,v_3,v_4,v_5,$ and $v_k$ are not linearly independent. Now, we see that for all $k=1,2,3,4,5$ $\left|M_k\right|$ is a sum of products $$v_{1,j_1}v_{2,j_2}v_{3,j_3}v_{4,j_4}v_{5,j_5}$$for $j_1,j_2,j_3,j_4,J_5\in\{1,2,3,4,5,6\}$, and as $v_{k,j}$ has degree at most $2d_k$ for all $k=1,2,3,4,5$ and $j=1,2,3,4,5,6$ we see that $$v_{1,j_1}v_{2,j_2}v_{3,j_3}v_{4,j_4}v_{5,j_5}$$has degree at most $2\left(d_1+d_2+d_3+d_4+d_5\right)$ for all $j_1,j_2,j_3,j_4,j_5\in\{1,2,3,4,5,6\}$, meaning that $\left|M_k\right|$ has degree at most $2\left(d_1+d_2+d_3+d_4+d_5\right)$ for all $k=1,2,3,4,5$. Now, we will consider any $t=t_0$ for which at least $2$ of $P_1,P_2,P_3,P_4,$ and $P_5$ are the same. We will show that $\left|M_k\right|=0$ when $t=t_0$ so that $\left|M_k\right|$ has a factor of $t-t_0$ for $k=1,2,3,4,5,6$, so that we can divide all coordinates of $v$ be $t-t_0$ and thus decrease the degree of each coordinate by $1$, which would prove the claim, similar to the proof of Zack's lemma. Without loss of generality, let $P_1=P_2$. Then, we see that $$\begin{pmatrix}x_1\\x_2\\x_3\end{pmatrix}=c\begin{pmatrix}x_2\\y_2\\z_2\end{pmatrix}$$for some $c$, so that $$v_1=c^2v_2.$$This implies that the topmost row vector of $M_k$ is exactly $c^2$ times the second topmost row vector, so $\left|M_k\right|=0$ for $k=1,2,3,4,5,6$ since then the row vectors of $M_k$ are not linearly independent, proving the claim. Now, we define a projective version of the power of a point as follows. For a circle with equation $$Ax^2+By^2+Cz^2+Dyz+Ezx+Fxy=0,$$we have that $A=B$ and $F=0$ as the circle has equation $$(x-az)^2+(y-bz)^2=(cz)^2$$for some $a,b,$ and $c$ which expanded has the $x^2$ and $y^2$ coefficient equal to $0$, so that the circle has equation $$Ax^2+Ay^2+Cz^2+Dyz+Ezx=0,$$a point $P=(x:y:z)$ has power $$\left(Ax^2+Ay^2+Cz^2+Dyz+Ezx:Az^2\right)\in\mathbb{CP}^1.$$This does not depend on the equation or $x,y,$ and $z$ given the circle and $P$. Claim. This matches with the ordinary definition of the power of a point for all instances where the ordinary definition is defined. Proof. Then, we have that since expanding $(x-az)^2+(y-bz)^2-(cz)^2=0$ for some $a,b,$ and $c$ can give an equation $Ax^2+Ay^2+Cz^2+Dyz+Ezx=0$ of the circle with $A=1$ we have that \begin{align*} \frac{Ax^2+Ay^2+Cz^2+Dyz+Ezx}{Az^2}^2=\frac{(x-az)^2+(y-bz)^2-(cz)^2}{1\cdot{}z^2}\\ &=\left(\frac{x}{z}-a\right)^2+\left(\frac{y}{z}-b\right)^2-c^2 \end{align*}which matches the ordinary definition. Lemma $2$. We can now define the degree of a power of a point in terms of a single variable $t$ as the minimum of the maximum degree of $A(t)$ and $B(t)$ over all polynomials $A(t)$ and $B(t)$ with the power being $(A(t):B(t))$. Then, given a fixed point $P$ and a circle $\omega$ moving with respect to $t$ with degree $d$ so that there are $x$ values of $t$ for which the circle degenerates into a line union the line at infinity so that $P$ is on the circle, the degree of the power of $P$ with respect to $\omega$ is at most $d-x$. Proof. Let $A,C,D,$ and $E$ vary with degree at most $d$ so that $\omega$ has equation $$Ax^2+Ay^2+Cz^2+Dyz+Ezx=0.$$Let $P=(p:q:r)$. Then, the power of $P$ with respect to $\omega$ can be expressed as $$\left(Ap^2+Aq^2+Cr^2+Dqr+Erp:Ar^2\right).$$Since $p^2,q^2,r^2,qr,rp,$ and $r^2$ are fixed, we see that both coordinates have degree at most $d$. Now, when the circle degenerates into a line union the line at infinity so that $P$ is on the circle, we see that $$Ap^2+Aq^2+Cr^2+Dqr+Erp=0$$since $P$ is on the circle and $$Ar^2=0$$since $A=0$ since the circle is a line union the line at infinity, so if this occurs when $t=t_0$ then we can divide both coordinates of the power of $P$ with respect to $\omega$ by $t-t_0$ since $$Ap^2+Aq^2+Cr^2+Dqr+Erp$$and $$Ar^2$$both have factors of $t-t_0$ since they are both $0$ when $t=t_0$. Performing this for all possible $t_0$ gives that the power of $P$ with respect to $\omega$ has degree at most $d-x$. [asy][asy] size(300); pair A,B,U1,U2,V1,V2,A1,B1,X,K,C,D; A=dir(105); B=dir(285); U1=dir(45); U2=dir(315); V1=dir(195); V2=dir(225); A1=U1*V1/A; B1=U2*V2/B; X=dir(210); K=extension(A1,B,A,B1); C=extension(A,X,U1,V1); D=extension(B,X,U2,V2); draw(Circle((0,0),1)); draw(U1--V1); draw(U2--V2); draw(B1--A--X--B--A1); draw(circumcircle(X,C,D)); dot("$A$",A,dir(A)); dot("$B$",B,dir(B)); dot("$U_1$",U1,dir(U1)); dot("$U_2$",U2,dir(U2)); dot("$V_1$",V1,dir(V1)); dot("$V_2$",V2,dir(V2)); dot("$A'$",A1,dir(A1)); dot("$B'$",B1,dir(B1)); dot("$X$",X,dir(X)); dot("$C$",C,dir(dir(V1-C)+dir(A-C))); dot("$D$",D,dir(dir(V2-D)+dir(B-D))); dot("$K$",K,dir(dir(A-K)+dir(B-K))); [/asy][/asy] Let $A'$ other than $A$ be such that $\overline{AA'}\parallel\overline{U_1V_1}$ and let $B'$ other than $B$ be such that $\overline{BB'}\parallel\overline{U_2V_2}$. We claim that we can have that $K=\overline{A'B}\cap\overline{AB'}$. Note that $OK^2-\rho^2$ is the power of $K$ with respect to $(XCD)$. Vary $X$ with degree $2$ on $\Gamma$. It suffices to show that the power of $K$ with respect to $(XCD)$ has degree $0$. We see that $\overline{AX}$ has degree at most $0+2-1=1$ by Zack's lemma since $X$ can equal $A$, so $C=\overline{AX}\cap\overline{U_1V_1}$ has degree at most $1+0=1$. Similarly, we see that $D$ has degree at most $1$. Now, note that when $X=U_1,V_1$ we have that $X=C$, when $X=U_2,V_2$ we have that $X=D$, and we can also have that $X=I,J$ since $I$ and $J$ are on $\Gamma$. This implies that $(XCD)$, the conic through $X,C,D,I,$ and $J$, has degree at most $$2(2+1+1+0+0)-6=2$$by lemma $1$ since we found $6$ cases where $2$ of $X,C,D,I,$ and $J$ are equal. Now, since when $X=A'$ we see that $(XCD)$ degenerates to $\overline{A'B}$ union the line at infinity and when $X=B'$ we see that $(XCD)$ degenerates to $\overline{AB'}$ union the line at infinity, the power of $K$ with respect to $(XCD)$ has degree at most $2-2=0$ since $K$ lies on both $\overline{A'B}$ and $\overline{A'B}$, so we are done.