How to prove that the tangent to a circle is perpendicular to the radius drawn to the point of contact? I've tried drawing a parallel chord to the tangent but then how would you prove that the chord is perpendicular to the radius?
 A: Suppose $\Gamma$ is a circle centered at $O$. Let $\ell$ be a line tangent to $\Gamma$ at a point $A$. Suppose the line from $O$ perpendicular to $\ell$ meets $\ell$ at a point $B$. If $B\neq A$, then there exists a point $C$ on $\ell$ on the other side of $B$ from $A$ such that $AB\cong BC$. (This follows from Hilbert's first axiom of congruence for line segments.)
By the side-angle-side theorem, $\triangle OBA\cong\triangle OBC$, and thus $OA\cong OC$. Thus $C\in\Gamma$. But $C\neq A$, a contradiction, since the tangent line $\ell$ can only meet $\Gamma$ at one point by definition of tangency. Thus $B=A$. Thus $\ell\perp OA$.
A: 
By using derivatives, slope of tangent at any point is -x/y. Slope of radius is y/x, which is negative reciprocal of slope of tangent. Therefore, angle OAQ is 90 degrees, because, straight lines whose slopes are negative reciprocals of each other are perpendicular
A: Let $O$ be the centre of the circle, let $\ell$ be a tangent line, and let $P$ be the point of tangency.  Suppose that $OP$ is not perpendicular to $\ell$. Draw the line through $O$ which is perpendicular to $\ell$. Then this line meets $\ell$ at a point $Q\ne P$.
Note that $Q$ is outside the circle. Now consider the triangle $OQP$. This is right-angled at $Q$.  So $OP$ is the hypotenuse of this triangle, and is therefore bigger than $OQ$. But this is impossible: since $Q$ is outside the circle, we must have $OP\lt OQ$. 
A: The following shows how a chord ( between separate real root intersections  ) become a single real repeated root at tangent point.
Let the circle be $ (x-h)^2 + y^2 = a^2 $. Let it be cut by a straight line through the origin $ y = m x $. There are two roots. For point of tangency there should be a double root or a single coincident root.
Let  $ h^2 = a^2 + T^2, $
$$ x^2(1+m^2) - 2 x h + T^2 = 0 $$
Discriminant $\Delta$ should vanish.
$$ \Delta =0, h^2 = T^2 (1+m^2) $$
The Pythagoras theorem is satisfied with $h$ as hypotenuse, $a$ radius and $T$ length of tangent, sides $a$ and $T$ are perpendicular with:
$$ \tan(\alpha) = m  = a/T , \sin(\alpha) = a/h $$

A: Basically, since the tangent is perpendicular to the radius drawn to the point of contact and the chord is parallel to the tangent, thus the chord is perpendicular to the radius drawn to the point of contact.
A: Extend the radius to a line $L$ and use reflection symmetry with respect to this line.
