Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I've tried drawing a parallel chord to the tangent but then how would you prove that the chord is perpendicular to the radius?

share|improve this question

3 Answers 3

up vote 1 down vote accepted

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 $OP\lt OQ$.

share|improve this answer

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 $\ell\perp OA$.

share|improve this answer

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.

share|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.