1. Solve the differential equation

Given $\frac{dy}{dx}+\frac{x+a}{y-2}=0$ so $\frac{dy}{dx}=-\frac{x+a}{y-2}.$

Rearrange: $(y-2)\,dy=-(x+a)\,dx.$

Integrating, $\int (y-2)\,dy = -\int (x+a)\,dx$ $\frac{y^2}{2}-2y=-\frac{x^2}{2}-ax+C.$

Multiply by $2$: $y^2-4y = -x^2-2ax+C_1.$

Hence $x^2+y^2+2ax-4y=C_2.$

Use the condition $y(1)=0$: $1+0+2a(1)-0=C_2 \implies C_2=1+2a.$

Therefore the curve is $x^2+y^2+2ax-4y=1+2a.$

Complete squares: $(x+a)^2+(y-2)^2 = a^2+5+2a=(a+1)^2+4.$

So $C$ is a circle with center $(-a,2)$ and radius $r=\sqrt{(a+1)^2+4}.$

2. Use the given area

Area enclosed is $4\pi$, so $\pi r^2=4\pi \implies r^2=4 \implies r=2.$

Thus $(a+1)^2+4=4 \implies (a+1)^2=0 \implies a=-1.$

Hence the circle is $(x-1)^2+(y-2)^2=4.$

3. Find points of intersection with the $y$-axis

On the $y$-axis, $x=0$. Then $(0-1)^2+(y-2)^2=4$ $1+(y-2)^2=4$ $(y-2)^2=3$ $y=2\pm \sqrt{3}.$

So the points are $P=(0,2+\sqrt{3}), \qquad Q=(0,2-\sqrt{3}).$

4. Find slope of tangent and normal

From the differential equation, with $a=-1$, $\frac{dy}{dx}=-\frac{x-1}{y-2}.$

At $P=(0,2+\sqrt{3})$

$m_t=-\frac{0-1}{\sqrt{3}}=\frac{1}{\sqrt{3}}.$ So slope of normal is $m_n=-\sqrt{3}.$

Equation of normal at $P$: $y-(2+\sqrt{3})=-\sqrt{3}(x-0).$ To find its intersection $R$ with the $x$-axis, put $y=0$: $-(2+\sqrt{3})=-\sqrt{3}x$ $x_R=\frac{2+\sqrt{3}}{\sqrt{3}}=1+\frac{2}{\sqrt{3}}.$

At $Q=(0,2-\sqrt{3})$

$m_t=-\frac{0-1}{-\sqrt{3}}=-\frac{1}{\sqrt{3}}.$ So slope of normal is $m_n=\sqrt{3}.$

Equation of normal at $Q$: $y-(2-\sqrt{3})=\sqrt{3}(x-0).$ Set $y=0$ for intersection $S$ with $x$-axis: $-(2-\sqrt{3})=\sqrt{3}x$ $x_S=\frac{\sqrt{3}-2}{\sqrt{3}}=1-\frac{2}{\sqrt{3}}.$

5. Compute length $RS$

Both $R$ and $S$ lie on the $x$-axis, so $RS=|x_R-x_S|.$

Thus $RS=\left(1+\frac{2}{\sqrt{3}}\right)-\left(1-\frac{2}{\sqrt{3}}\right)=\frac{4}{\sqrt{3}}=\frac{4\sqrt{3}}{3}.$

6. Match with options

$RS=\frac{4\sqrt{3}}{3}.$ So the correct option is A.