الأسبوع 8 · تطبيقات التشابه في الدائرة: الوتران المتقاطعان والقاطعان والمماس وعكس ذلك · 3 دروس
2
A tangent and a secant from one outside point: (EA)² = ED × EC
Mathematics Sec 1 Term 1, Unit 2, Lesson 4
GoalUse the tangent-secant relation (EA)² = ED × EC.
New words
tangent segment · the segment from the outside point to the point where the tangent touches the circleouter part · the part of the secant from E to the first point D
Explanation
Take a point E outside the circle. A tangent EA touches the circle at A, and a secant from E meets the circle at D and then at C. Then (EA)² = ED × EC. This is the secant theorem when the two points of one secant come together: the second secant turns into a tangent, and EA × EA = EC × ED. Check with a circle of radius 3 and a point E at distance 5 from the centre O. The radius OA is perpendicular to the tangent, so by Pythagoras EA² = 5² − 3² = 16 and EA = 4. The secant through the centre meets the circle at distances 5 − 3 = 2 and 5 + 3 = 8 from E, and ED × EC = 2 × 8 = 16 ✓. In every case use the whole secant EC and the outer part ED, and never EA = ED × EC (the tangent is squared). Two tangents from the same point have equal lengths.
Worked examples
EA is a tangent, ED = 2 and DC = 6. EC = ED + DC = 8, so EA² = ED × EC = 2 × 8 = 16 and EA = 4 cm.
EA = 10 is a tangent and DC = 21. Let ED = x. Then x(x + 21) = 10² = 100, x² + 21x − 100 = 0, (x − 4)(x + 25) = 0, so x = 4. ED = 4 and EC = 25 (check 4 × 25 = 100 ✓).
A tangent has length 12 and the outer part of a secant is 8. Then EC = 144/8 = 18 and DC = 18 − 8 = 10.
Class activity
Two tangents. Draw a circle, mark a point E outside, and draw the two tangents from E with a set square. Measure both tangent lengths. What do you notice?
Practice
1
EA is a tangent, ED = 4 and EC = 16. Find EA and DC.
