Tangents from a common external point are equal, and the line to the centre bisects the angle.
Key formulas
Equal lengths
From P, tangents touch at T and Q. Then PT=QT. Triangles OPT and OQZ are congruent (RHS or hypotenuse-leg style with radii).
Angles
OP bisects ∠TPQ and also bisects ∠TOQ. Radii OT, OQ are perpendicular to the tangents.
Example
PT=12. Then QT=12. If ∠TPQ=40°, each half is 20°.
Worked example
Worked example
Problem. From external point P, two tangents touch a circle at T and Q. PT = 12 cm and ∠TPQ = 40°. Find QT and ∠TPO if O is the centre and OP bisects ∠TPQ.
Solution. Tangents from an external point are equal: QT = 12 cm. OP bisects ∠TPQ, so ∠TPO = 20°.
Kite
Quadrilateral OT PQ often forms a kite with right angles at T and Q.
Review quiz
From P, PT=QT=9 and radius OT=12. Distance OP is:
Practice questions
- Tangents from P are each 9 cm. What else is equal?
- Radius 5, tangent length 12. Find the distance OP.
Study notes
Distance OP?
OP² = OT² + PT² by Pythagoras.
More in Plane Geometry
- Circle tangents: the radius is perpendicular
- Right-triangle lengths: Pythagoras and altitude
- Similar triangles: matching sides in proportion
- Chord theorem: perpendicular from the centre
- Isosceles triangles and equal base angles
- Alternate segment theorem with a short proof
- Converse of Pythagoras: proving a right angle
- Heron’s formula when height is hidden