When two tangent segments are drawn from the same external point to a circle, the segments have equal length. This theorem is useful because it lets you find missing lengths without measuring the circle directly. In a diagram with external point P and tangent points A and B, the result is PA = PB.
The equal tangents also create a symmetric kite-shaped quadrilateral when the radii OA and OB are drawn.
Key Facts
- If PA and PB are tangents from the same external point P, then PA = PB.
- A radius to a point of tangency is perpendicular to the tangent line, so OA ⊥ PA and OB ⊥ PB.
- The quadrilateral OAPB is a kite because OA = OB and PA = PB.
- Right triangles OAP and OBP are congruent by HL because OP is shared and OA = OB.
- Tangent length formula from external point: PA = sqrt(OP^2 - r^2), where r is the circle radius.
- If PA = 3x + 2 and PB = 5x - 8, then set 3x + 2 = 5x - 8 to solve.
Vocabulary
- Tangent
- A tangent is a line or segment that touches a circle at exactly one point.
- Point of tangency
- The point of tangency is the single point where a tangent touches a circle.
- External point
- An external point is a point located outside a circle.
- Radius
- A radius is a segment from the center of a circle to any point on the circle.
- Kite
- A kite is a quadrilateral with two pairs of adjacent equal sides.
Common Mistakes to Avoid
- Setting the tangent segments unequal, such as PA > PB, is wrong because tangents drawn from the same external point are congruent.
- Forgetting that the radius is perpendicular to the tangent is wrong because the right angle is the key fact that allows right-triangle reasoning.
- Using the diameter instead of the radius in PA = sqrt(OP^2 - r^2) is wrong because the right triangle uses OA or OB as one leg, and that length is the radius.
- Assuming any two segments from P to the circle are equal is wrong because the theorem applies only to tangent segments, not secants or chords.
Practice Questions
- 1 From external point P, tangents PA and PB touch a circle at A and B. If PA = 14 cm, what is PB?
- 2 A circle has radius 6 cm and OP = 10 cm. If PA is tangent from P to the circle, find PA using PA = sqrt(OP^2 - r^2).
- 3 Explain why quadrilateral OAPB forms a kite when PA and PB are tangents from P and O is the center of the circle.