335. The Tangent's Promise · always at right angles to the radius

Try to break it

Drag P around the circle. The tangent at P always meets the radius OP at exactly 90°. Try to tilt the tangent off that 90°; the construction won't allow it. If it tilted, the line would cut the circle a second time — and a tangent touches the circle at exactly one point.

How you build it

Mark the centre O, draw the circle, mark the point of contact P, then build the tangent at P perpendicular to the radius OP.

• Mark the centre O of the circle.
• With centre O, draw a circle of radius about 200.
• Place a point P anywhere on the circumference.
• Draw the radius OP connecting the centre O to P.
• Construct the tangent line at P, perpendicular to OP.

The proof, step by step

Prove that the radius to the point of contact is perpendicular to the tangent.

• Take any point Q on the tangent AB, other than P.
• Join OQ. In triangle OPQ, angle OPQ is 90°, so OQ is the hypotenuse.
• The hypotenuse is always the longest side, so OQ > OP.
• Since OP is the shortest segment from O to the tangent line, OP must be perpendicular to AB.

Worked example

In the given figure, AB is a tangent to a circle with centre O at point P. If OP = 7 cm and PQ = 24 cm (where Q is on AB), what is the length of OQ?

Since OP ⊥ AB, triangle OPQ is right-angled at P. By Pythagoras theorem, OQ² = OP² + PQ² = 7² + 24² = 49 + 576 = 625. Thus, OQ = 25 cm.

• 17 cm
• 25 cm — correct
• 31 cm
• 20 cm