Question is: Find the point on the circle $x^2+y^2=1$ closest to the line given by $y=4-\sqrt{3}x$ using multivariable calculus.
I know the shortest distance is the distance between the center and its orthogonal projection on the line minus the radius.
So this could be translated to(i think):
$\vert \frac{\sqrt{3}x+y-4}{2} \vert$, and then i have to study the critical points of this function, is this correct? and how do i get rid of the absolute value?
I cannot use Lagrange Multipliers either.
By gradients we need to set that they are parallel that is
$$(\sqrt 3,1)=\lambda(2x,2y)\implies x=\sqrt 3 y$$
and by equality of the tangents we have
$$\frac{dy}{dx}=-\frac{x}{y}=-\sqrt 3\implies x=\sqrt 3 y$$
thus in both cases
$$x^2+y^2=1 \implies 3y^2+y^2=1\implies y=\pm \frac12 \quad x=\pm \frac {\sqrt 3} 2$$
and the solution is $P=\left(\frac{\sqrt 3}2,\frac12\right)$.