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 $a+\sqrt 3 b=0\implies b=0\implies x=1,y=\sqrt 3,z=c=\pm1$
Call your curve $c$, then $t=1$ and the tangent line is achieved as usual:
$$s\mapsto c(t)+s\cdot \dot c(t)=(3,0,2)+s(1,3,3).
$$
Michael
Best Answer
Assuming you are right about knowing how to find $T(t)$( I believe in you, you can do it!) Recall that a line can be parameterized as $l(t)=v_0+tv$ where $v_0$ is a position point(i.e. the value of your function at the point you want to find the tangent) and $v$ the direction vector(i.e. the "slope" of your line, so the tangent to your curve at the desired point.)