The equation of any line passing through $(1,\sqrt 3,0)$ can be written as
$\frac{x-1}a=\frac{y-\sqrt 3}b=\frac z c$ where $a^2+b^2+c^2=1$
So,$cx=az+c,cy=bz+\sqrt 3c$
Putting the values of $x,y$ in $x^2+y^2=4,$
$(az+c)^2+(bz+\sqrt 3c)^2=4c^2$
$(a^2+b^2)z^2+2zc(a+\sqrt 3 b)=0$
But this is a quadratic in $z,$, each root represent the $z$ co-ordinate of the intersection.
For tangency, both root should be same, so $\{2c(a+\sqrt 3 b)\}^2=4(a^2+b^2)0$
$\implies c(a+\sqrt 3 b)=0--->(1)$
Putting the value of $x$ in $x^2+z^2=1,$
$(az+c)^2+c^2z^2=c^2\implies (a^2+c^2)z^2+2caz=0$
So like 1st case, $(2ca)^2=4(a^2+c^2)0\implies ca=0--->(2)$
Form $(1)$ and $(2)$,
if $c=0,a^2+b^2=1\implies x=a+1,y=b+\sqrt 3,z=0 $
if $c\ne 0,a=0$ and $a+\sqrt 3 b=0\implies b=0\implies x=1,y=\sqrt 3,z=c=\pm1$
We wish to find where our parameterized curve intersects the line $2y+x=7$. Let's parameterize that line by $x=2t+3,$ so we'll need $y=-t+2$. Under this parameterization, the $x$-values of our line and curve will be the same at any $t$, so we need only determine where the $y$-values are the same--i.e. find $t$ for which $$t^3-4t=-t+2\\t^3-3t-2=0.$$ Recall, though, that we already know they meet where $t=-1$, so we know we have a factor of $t+1$ in our cubic (that's why we went with this method, instead of yours). In particular, $$t^3-3t-2=(t+1)(t^2-t-2)=(t+1)^2(t-2),$$ so the cutting point occurs when $t=2$.
Best Answer
Suppose your parametric curve is $\mathbf{C}(t) = (x(t), y(t))$ and your circle has its center at $\mathbf{Q} = (a,b)$ and radius $r$.
Then, the equation of the circle is $\|\mathbf{X} - \mathbf{Q}\| = r$. So, at intersection points $$ \|\mathbf{C}(t) - \mathbf{Q}\| = r $$ Working with coordinates, instead, the equation of the circle is $(x-a)^2 + (y-b)^2 = r^2$, so, at intersection points, we have $$ (x(t)-a)^2 + (y(t)-b)^2 = r^2 $$ Saying it another way, you need to find the zeros of the function $$ f(t) = (x(t)-a)^2 + (y(t)-b)^2 - r^2 $$ You'll need to use numerical methods to do this, typically. If you use a Newton-Raphson style of root finder, then the derivatives you mentioned will be handy.
In the special case where $\mathbf{C}(t)$ is a polynomial of degree 3 (a Bezier curve, for example), then $f(t)$ will be a polynomial equation of degree 6, so you'll still have to use numerical methods to find solutions.
The other answer you cited is basically correct. As you guessed, $(x_C,y_C)$ denotes the center of the circle. But the right-hand side of the equation he gives should be $r^2$, not $0$.