Nice problem but not hard. Assume that C=(O,R),C'=(O_1,R_1) and C"=(O_2,R_2).OX=OY=R,O_1X=O_1Z=R_1 and O_2Y=O_2Z=R_2 => circumcircle of XYZ it is incircle of OO_1O_2.I am prove that BX prpendicular of OO_1 and AY prpendicular of OO_2=>Q is the circumcenter of XYZ.AY prpendicular of OO_2 <=>AO^2-(AO_2)^2=OY^2-(O_2Y)^2=R^2-(R_2)^2,AO^2=R^2+4RR_1 and (AO_2)^2=(2(RR_1)^1/2-2(RR_2)^1/2)^2+(2R-R_2)^2=>AO^2-(AO_2)^2=R^2-R_2^2<=> R^2=4R_1R_2 but this is easy question.

In addition, I proved that: ${\overline {\underline |\ OZ\perp AB\ | }}$ (very nice !); $\ 2\rho=\frac{R\sqrt R}{\sqrt R_1 +\sqrt R_2}$, where $\rho$ is the inradius of the triangle $XYZ$ (easily); $OZ\perp O_1O_2\Longleftrightarrow R_1=R_2$ (easily).

Let $C$ be tangent to $L'$ at $D$. A homothety centered at $Y$ maps $C$ to $C''$. Since $L'$ and $L''$ are parallel, $D$ is mapped to $B$, so $D$, $Y$, and $B$ are collinear. Similarly, $A$, $Z$, and $B$ are collinear. Let the common external tangents of $C$ and $C'$ intersect at $P$. is the external homothety center of $C$ and $C'$. $Y$ and $Z$ are the internal homothety centers of $C$ and $C''$ and $C'$ and $C''$, respectively. Hence, by d'Alembert's theorem, $P$, $Y$, and $Z$ are collinear. Let $PZ$ intersect $C'$ at $Y'$. Since the homothety centered at $P$ maps $C'$ to $C$, $\frac{PY'}{PA} = \frac{PY}{PD}$, so $PY' \cdot PD = PA \cdot PY$. By power of a point on $C'$, $PA^2 = PY' \cdot PZ$, so $PY' = \frac{PA^2}{PZ}$. Substituting this into $PY' \cdot PD = PA \cdot PY$, so $ADYZ$ is cyclic. Since $B = AZ \cap DY$, $BY \cdot BD = BZ \cdot BA$, so $B$ lies on the radical axis of $C$ and $C'$. By symmetry, $A$ must lie on the radical axis of $C$ and $C'$, so $Q$ must be the radical center of $C$, $C'$, and $C''$. Since $QX$, $QY$, and $QZ$ are tangents and $Q$ has equal power with respect to each of these circles, $QX = QY = QZ$, as desired.

Strangely enough, I found G3 harder than G4. Let $C$ touch $L'$ and $L''$ at $M$ and $N$ repectively. Let the centre of $C$ be $O_1$, the centre of $C'$ be $O_2$ and centre of $C''=O_3$. Join $O_1,X$ and $O_2,X$ as $AO_2\parallel MN\implies \angle AO_2X=\angle XO_1N$ $\implies \Delta XO_1N~\Delta AO_2X\implies \angle AXO_2~\angle O_1XN\implies A,X,N$ are collinear. Similarly we can prove that $A,Z,B$ are collinear and $B,Y,M$ are collinear. $\angle AZX=\angle MAN$(alternate segment, $L'$ is tangent to $C'$) $\angle MAN=\angle ANM$ as $C$ is tangent to $L'$ and $L''$ at diametrically opposite points. $\implies \angle AZX=\angle XNB\implies XZMN$ is cyclic, $XN\cap ZB=A\implies A$ lies on the radical axis of the circles $C$ and $C''$, but as $C$ and $C''$ touch each other $\implies AY$ is the direct common tangent to $C$ and $C''$. Similarly, we can prove that $BX$ is the direct common tangent to $C$ and $C'$. $\implies AY\perp O_1O_3$ and $BY\perp O_1O_2$ $X,Y,Z$ are the points where the incircle of $\Delta O_1O_2O_3$ touches the sides. This means that the circumcentre of $\Delta XYZ=$ the incentre of $\Delta O_1O_2O_3$ As $AY\perp O_1O_3\implies AY$ passes through the incentre of $\Delta O_1O_2O_3$ $BX\perp O_1O_2\implies BX$ passes through the incentre of $\Delta O_1O_2O_3$ $\implies AY\cap BX=\text{ the incentre of } \Delta O_1O_2O_3=\text{the circumcenter of} \Delta XYZ$