341. Tangents from the Outside · equal lengths, perfect right angles
PA and PB are always equal in length, and both meet the circle at exactly 90°.
What this lesson covers
Try to break it
Drag P around outside the circle. The two tangent lines from P always come out equal in length (PA = PB), and each meets its radius at 90°. Pull P inside the circle and the construction can't draw tangents — no line through an interior point can touch the circle at exactly one place.
How you build it
Construct the two tangents from an external point P, and see that PA = PB with each meeting the circle at 90°.
- Point tool: mark O near the middle — the centre of the circle.
- Circle tool: click O as the centre, then click outward to set the radius.
- Point tool: mark P well outside the circle — the external point.
- Segment tool: join the centre O to the external point P.
- Bisector tool: click O then P — it draws the perpendicular bisector of OP, crossing OP at its midpoint.
- Point tool: mark M where the perpendicular bisector crosses OP — the midpoint of OP.
- Circle tool: click M as the centre, then click O — this circle has OP as its diameter and cuts the main circle at the points of contact.
- Point tool: mark A at one point where the construction circle meets the main circle.
- Point tool: mark B at the other crossing of the two circles.
- Segment tool: join P to A — the first tangent.
- Segment tool: join P to B — the second tangent. Because A and B lie on the circle with diameter OP, ∠OAP = ∠OBP = 90°, so PA and PB are tangents — and PA = PB.
The proof, step by step
Prove that the two tangents constructed from the external point are equal and meet the circle at right angles to the radius.
- A and B lie on the circle with diameter OP.
- ∠PAO = 90° and ∠PBO = 90° (Angles in a semicircle).
- OA ⊥ PA and OB ⊥ PB.
- PA and PB are tangents (perpendicular to radius at endpoint).
Worked example
From an external point P, a tangent PT is drawn to a circle with centre O and radius 5 cm. If OP = 13 cm, find the length of PT.
In right triangle OTP (right-angled at T), PT² = OP² - OT² = 13² - 5² = 169 - 25 = 144. Thus, PT = √144 = 12 cm.
- 12 cm — correct
- 8 cm
- 10 cm
- 13 cm