Find all $(x,y,z)$ on $S$ such that the line connecting $Q$ to $(x,y,z)$ is tangent to $S$? Part A: Let $C$ be the unit circle in $\mathbb{R}^2$ and let $P$ be the point $(3,0)$. Find all $(x,y)$ on $C$ such that the line intersecting $P$ to $(x,y)$ is tangent to $C$?
Can someone help me with Part A so that I have an idea of what to do. Then I will give Part B a shot on my own. Part B is the title. 
 A: Since in my calculus days the unit circle was always centered at the origin I'll give another (different) answer from this perspective:
The tangent line to the curve $x^2+y^2=1$ at $(x,y)$ has slope $m = -\dfrac{x}{y}$, which can be found with implicit differentiation. 
A line passing through $(x,y)$ and $(3,0)$ has slope $m=\dfrac{y}{x-3}$. Setting these equal:
$$\dfrac{y}{x-3} = \dfrac{-x}{y} \iff \left(x-\dfrac{3}{2}\right)^2+y^2=\dfrac{9}{4}$$
While not necessary, I wrote in this form, because it will be the case that the tangent lines pass through the intersections of two circles. So all points $(x,y)$ for which a line through it and $(3,0)$ belongs to both this and the unit circle. Subtracting the equation of the first circle from the second we have
$$\left(x-\dfrac{3}{2}\right)^2-x^2=\dfrac{5}{4}$$
Solving for $x$ gives $x=\dfrac{1}{3}$ and hence $y=\pm\dfrac{2\sqrt{2}}{3}$. The points on the circle are then
$$\left(\dfrac{1}{3},\dfrac{2\sqrt{2}}{3}\right),\ \ \ \ \left(\dfrac{1}{3},-\dfrac{2\sqrt{2}}{3}\right)$$
So the lines are
$$y=\dfrac{-1}{2\sqrt{2}}(x-3)$$
$$y=\dfrac{1}{2\sqrt{2}}(x-3)$$
See here: https://www.desmos.com/calculator/h0m1phgliu
A: Any point on a unit circle is given in terms of its center $(x_0, y_0)$ and a parameter $\theta$:
$$
x = x_0 + cos\theta   \\
y = y_0 + sin\theta
$$
If such a point $(x, y)$ is a point of tangency from the point $(3,0)$ then the corresponding line segment from $(x_0,y_0)$ to $(x,y)$ is perpendicular to the one from $(3,0)$ to $(x,y)$ and so the corresponding dot product is zero:
$$
(x-3)(x - x_0) + (y - 0)(y - y_0) = 0 \\
(x_0 + cos\theta - 3)cos\theta + (y_0 + sin\theta)sin\theta = 0
$$
Solving for $\theta$ in principle gives you the required value(s) of $(x,y)$. 
