What kind of a locus $Y$ describe when moving $\Omega $? Two circles $\Gamma_1$ and $\Gamma_2$ have common external tangents $\ell_1$ and $\ell_2$ meeting at $T$. Suppose $\ell_1$ touches $\Gamma_1$ at $A$ and $\ell_2$ touches $\Gamma_2$ at $B$. Variable circle $\Omega$ through $A$ and $B$ intersects $\Gamma_1$ again at $C$ and $\Gamma_2$ again at $D$, such that quadrilateral $ABCD$ is convex.
Suppose lines $AD$ and $BC$ meet at point $Y$.
What kind of a locus $Y$ describe when moving $\Omega $? Is that  some projective curve?

How can we recognize it with pure Euclidian geometry without any (or as few as possible) anaylitic calculation? For example, if we have a projective transformation $\pi$ between pencils of lines, then we know that $p\cap \pi(p)$ describes some conic or line.
 A: It's not clear how OP expects to "recognize [the locus] with pure Euclidian geometry". Cubic curves come in many shapes. So I cheated and used coordinates (and some symbol-crunching in Mathematica). Perhaps the finding here can inspire a "pure" solution.

Taking $T$ at the origin, $\angle ATB=2\theta$, $a:=|TA|$, and $b:=|TB|$, I get this equation for the locus:
$$\begin{align}
0 &= (x^2+y^2)(a+b) \cos\theta \left(\;x (a-b) \sin\theta - y (a + b) \cos\theta\;\right) \\[4pt]
&\quad+ 2 x^2  a b (2y -(a-b)\sin\theta )\\[4pt]
\end{align} \tag1$$
As a polar curve parameterized by $\phi$, this is:
$$r = \frac{2 a b (a - b)\sin\theta\cos^2\phi}{
  2 a b \sin2\phi\cos\phi
   - (a + b) \cos\theta (a \sin(\phi-\theta) + b \sin(\phi+\theta))
  )} \tag2$$
In barycentric coordinates $\alpha:\beta:\tau$ relative to $\triangle ABT$:
$$4 a b (a+b) \,\alpha\beta(\alpha - \beta) \sin^2\theta = (a - b) (a\alpha + b \beta)^2 \tau \tag{3}$$
Here are figures, with and without circle overlap:



*

*Four points are common to all instances: $A$, $B$, $T$, and the midpoint of $\overline{AB}$ (labeled $M$).


*When the circles overlap, their intersections ($P$ and $Q$) are on the curve.


*The curve crosses the $x$-axis at $H := \left(\frac{2ab}{(a+b)\cos\theta},0\right)$. If $A_\star$ and $B_\star$ are the respective centers of the $A$ and $B$ circles, then $|TH|$ is the harmonic mean of $|TA_\star|$ and $|TB_\star|$.
Finding other landmarks and properties is left as an exercise to the reader.
A: It is well known that a conic can be generated by the pairwise intersections of two pencils of lines. This answer attempts to explain how the locus of $Y$ can be generated by the pairwise intersections of a pencil of lines and a pencil of conics.  Much of what is presented here is observational and practical so that the reader can attempt to reproduce the construction in, say, Geogebra.

*

*for some preliminary background and references, refer to this answer to the question Geometric interpretation of cubic curve?


*let $L=0$ and $M=0$ be two equations (in $x$ and $y$) of lines $\ell,m$.  Then $L+\lambda M=0$ is the equation of a pencil of lines through $P=\ell\cap m$, parameterized by $\lambda$.  Note that since the equation for a line is not unique, the parameterization of a pencil is also not unique. (e.g. $2L=0$ is also an equation for $\ell$ and we will say $2L\cong L$, meaning they represent the same curve)


*similarly, if $C=0$ and $D=0$ are equations for conics $c$ and $d$, $C+\lambda D=0$ is an equation of a pencil of conics.  If $c$ and $d$ intersect in four points, then this pencil is the set of all conics that pass through all four points (called base points in this context).  Again, the parameterization is not unique because  $sC=0$ is an equation for $c$ for any $s$ ($C\cong sC$).


*if we have two pencils as above
$$
\begin{aligned}
L+\lambda M&=0 \\
C+\lambda D&=0
\end{aligned}
$$
we can eliminate $\lambda$ to get $\mathcal{K}=LD-CM=0$, which is a cubic that is the locus of the points $(L+\lambda M)\cap(C+\lambda D).$ $\mathcal{K}$ will run through $P$ and the four base points.  A good beginner's introduction to this is Haskell, Special cubics and quartics produced by projective pencils of straight lines and conics (a Master's thesis dated 1916).


*suppose we have a sufficiently well behaved cubic $\mathcal{K}$.  How do we reverse engineer to get the line and conic pencils that generate it?  Briefly, we fix four points $P_1,\dots,P_4$ on $\mathcal K$ and find their coresidual point $P$ as explained in the math.SE answer and illustrated at Wilkin, The coresidual of four points on a cubic.  We then select three lines$L,M,N$ through $P$ and the corresponding conics $C,D,E$ running through the points $P_1,\dots,P_4$ such that $L\cap C \in \mathcal{K}$ (and similarly for the pairs $M,D$ and $N,E$. $E$ will be in the pencil generated by $C$ and $D$ in the sense that $E\cong C+\lambda_E D$ for some $\lambda_E D$.  It is not necessarily true that $N\cong L+\lambda_N M$.  But there will some $s$ so that $N\cong L+\lambda_N sM$, so we replace the line pencil with $L+\lambda s M$ so that lines and conics are paired correctly to produce $\mathcal K$.


*with these preliminaries we can turn to the scenario in the OP.  We borrow several new points from Blue's answer. $M$ is the midpoint of $AB$. $A_\star$ and $B_\star$ are the centers of the $A$ and $B$ circles.  $H$ is the harmonic conjugate of $T$ with respect to $A_\star,B_\star$. $\mathcal K$ is the locus of $Y$, which we know is a cubic that goes through $T,H,A,B,M$.


*we choose $T,H,A,B$ as base points for the conic pencil.  The base point for the line pencil is $T$, the coresidual point of the base points.  The reader is invited to experimentally confirm this in Geogebra.


*We now need to find three line-conic pairs that define the pencils we need.  By inspection (eyeballing) in Geogebra these are (they may be more, but I couldn't find any)

*

*$(L,C)$ - the line through $T$ perpendicular to $TH$ plus the conic whose tangent at $T$ is the aforementioned line. (the fact that the line and tangent are the same is the reason for the cusp at $T$).

*$(M,D)$ - the line $TM$ and the degenerate conic consisting of the lines $AB$ and $TH$.

*$(N,E)$ - the line $TH$ and the conic whose tangent at $H$ goes through $M$



*With these pairs we can define the pencils which I claim generate $\mathcal K$. I have not tried to prove this, but the most direct confirmation would probably be to show that $LD-CsM=0$ is equivalent to the Cartesian equation derived by Blue.


*I also have no conceptual explanation why this construction is equivalent to the original construction using $Y$ in the $OP$.  But cubics can have multiple characterizations, e.g. the Apollonian Strophoid has at least six.
Finally a crude working diagram of the construction.  The conics $C$ and $D$ are shown in thick blue dotted lines.  A line and conic are in red intersecting at G. The locus of intersection is a thick black dotted curve overlaying Blue's equation (in bright green).  Better diagram to come. (maybe)

