The square corner at the contact point

A tangent touches a circle, centre O, at P. How is the radius OP placed against it?

The rule

Two lines are perpendicular when they meet at a right angle, 90°.

The radius to the point of contact is perpendicular to the tangent.

The proof

Suppose OP is not perpendicular to the tangent. Draw the perpendicular from O to the tangent. It meets the tangent at R, not at P.

Now take P′ on the other side of R, with RP′ = RP.

Look at triangles ORP and ORP′. OR is shared. The angles at R are right angles. Also RP = RP′. Two sides and the angle between them are equal. So the triangles match, and OP′ = OP.

Then P′ is as far from O as P is. So P′ lies on the circle. P′ also lies on the tangent. That is two shared points. A tangent has only one.

So OP must be perpendicular to the tangent.

The other way round

Let P be a point on the circle. Draw a line through P at a right angle to OP. Take any other point Q on it. Then OQ² = OP² + PQ², by Pythagoras.

Q is not P, so PQ is more than 0. So OQ is longer than the radius OP. Every other point of the line is outside the circle. The line is a tangent.

A diameter has two ends. The tangent at each end is at 90° to the diameter. Two lines at 90° to one line are parallel. So the two tangents are parallel.

Tap a tilt. Find the tangent.