Prove that a circle has an infinite number of tangents

Question

It seems obvious that a circle is comprised of the set of all points that are equidistant from one point, and that each point on the circumference of the circle represents a tangent. This seems to indicate that there are an infinite number of tangents to each circle. I'm curious to know if there exists a way to prove this algebraically.

One thought I had would be to take a square (as an example of an inscribed quadrilateral) and remove the corners iteratively (that is, to identify the tangent of the circle perpendicular to the angular bisector of each angle of the polyhedron and remove the area that does not contain the circle(So, for example, a square becomes a tangential octagon, which then becomes a tangential hexdecagon, etc)). As the number of iterations approaches infinity, the difference in area between the circle and the polygon would approach zero, or put another way, the polygon's area will not be equal the the area of the circle until the number of iterations (and therefore the number of tangents) is equal to infinity.

Sorry for rambling.

My questions are these: is my logic sound, and how can I express this algebraically?

Answer 1

For $x\in[0,1]$, let $y=\sqrt{1-x^2}$. This is a quarter circle at the origin with radius $1$. The derivative is $\frac{dy}{dx} = \frac{-x}{\sqrt{1-x^2}}$, which can take all the values in $(-\infty,0]$ for $x\in[0,1]$. Hence, "the circle has infinitely many tangents."

Answer 2

Fix the circle with radius $r$ in the plane with its centre at the origin. A tangent to the circle is the unique line which is perpendicular to the arm extending from the origin at an angle $\theta$ from the $x$-axis that intersects that arm at a distance of $r$ from the origin. Hence there is a one-to-one mapping between the values in $[0,2\pi]$ and the tangents to the circle.