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?

Solution 1:

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 $OP\lt OQ$.

Solution 2:

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 $\ell\perp OA$.

Solution 3:

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