Question:

Step 1: Identify the equations of the given circles. The equations of the two circles are provided. We observe that both circles pass through the origin $(0,0)$ because they lack a constant term. The first circle, $S_1$, is given by: $$ x^2 + y^2 + (3 + \sin \beta)x + (2\cos \alpha)y = 0 $$ The second circle, $S_2$, is given by: $$ x^2 + y^2 + (2\cos \alpha)x + 2cy = 0 $$ Step 2: Apply the condition for circles touching at the origin. If two circles touch each other at the origin $(0,0)$, their tangents at the origin must be identical. For a general circle $x^2 + y^2 + Gx + Fy + K = 0$, if $K=0$ (i.e., it passes through the origin), the equation of the tangent at the origin is $Gx + Fy = 0$. Step 3: Find the equation of the tangent at the origin for the first circle. For the first circle, $S_1: x^2 + y^2 + (3 + \sin \beta)x + (2\cos \alpha)y = 0$, the coefficients of $x$ and $y$ are $G_1 = (3 + \sin \beta)$ and $F_1 = (2\cos \alpha)$, respectively. Thus, the equation of the tangent to $S_1$ at the origin is: $$ (3 + \sin \beta)x + (2\cos \alpha)y = 0 \quad \text{(Equation 1)} $$ Step 4: Find the equation of the tangent at the origin for the second circle. For the second circle, $S_2: x^2 + y^2 + (2\cos \alpha)x + 2cy = 0$, the coefficients of $x$ and $y$ are $G_2 = (2\cos \alpha)$ and $F_2 = 2c$, respectively. Thus, the equation of the tangent to $S_2$ at the origin is: $$ (2\cos \alpha)x + 2cy = 0 \quad \text{(Equation 2)} $$ Step 5: Equate the coefficients of the tangent lines. Since the two circles touch at the origin, their tangent lines at the origin must be identical. This implies that the coefficients of $x$ and $y$ in Equation 1 and Equation 2 must be proportional. $$ \frac{3 + \sin \beta}{2\cos \alpha} = \frac{2\cos \alpha}{2c} $$ Now, we solve for $c$: $$ (3 + \sin \beta)(2c) = (2\cos \alpha)(2\cos \alpha) $$ $$ 2c(3 + \sin \beta) = 4\cos^2 \alpha $$ $$ c = \frac{4\cos^2 \alpha}{2(3 + \sin \beta)} $$ $$ c = \frac{2\cos^2 \alpha}{3 + \sin \beta} $$ Step 6: Determine the maximum value of $c$. To maximize the value of $c$, we need to maximize the numerator and minimize the denominator. The maximum value of $\cos^2 \alpha$ is $1$, which occurs when $\cos \alpha = \pm 1$ (e.g., $\alpha = 0, \pi$). The minimum value of $3 + \sin \beta$ occurs when $\sin \beta$ is at its minimum value, which is $-1$. Substituting these values: $$ c_{\text{max}} = \frac{2(1)}{3 + (-1)} $$ $$ c_{\text{max}} = \frac{2}{2} $$ $$ c_{\text{max}} = 1 $$ The maximum value of $c$ is $1$. The final answer is $\boxed{1}$.