F5 Maths circle tangents

Anonymous

2011-01-27 1:28 am

Two tangents are drawn to the circle x^2 + y^2 - 6x + 6y - 2 = 0. If the two tangents are parallel to the straight line L: 2x+ y + 1 = 0, finda) the equations of the two tangents
b) the coordinates of the point of contace for each tangent found in (a)

更新1:

Could you give me more detailed steps from (4c + 18)^2 - 20(c^2 + 6c - 2) = 0 to c^2 - 6c - 91 = 0

更新2:

From x^2 + 2x + 1 = 0 to x = -1 then y = -5. So, contact point (-1, -5) How to know that x = -1 then y = -5 ?

更新3:

How to know x = -1 ? 我可能睇miss左D野, 希望您耐心指導^^

更新4:

http://hk.knowledge.yahoo.com/question/question?qid=7011012401441 唔該答埋呢到, 我會比哂最佳您

回答 (1)

myisland8132

2011-01-27 1:43 am

✔ 最佳答案

(a) The slope of the tangents = -2

Let the equation of the tangent is

y = -2x + c, Sub. into C: x^2 + y^2 - 6x + 6y - 2 = 0

x^2 + (c - 2x)^2 - 6x + 6(c - 2x) - 2 = 0

5x^2 - (4c + 18)x + (c^2 + 6c - 2) = 0

Discriminant = 0

(4c + 18)^2 - 20(c^2 + 6c - 2) = 0

c^2 - 6c - 91 = 0

(c - 13)(c + 7) = 0

c = -7 or c = 13

The equations of the tangents are

y = -2x - 7 and y = -2x + 13

(b) Consider 5x^2 - (4c + 18)x + (c^2 + 6c - 2) = 0

when c = -7

x^2 + 2x + 1 = 0

x = -1 then y = -5. So, contact point (-1, -5)

when c = 13

x^2 - 14x + 49 = 0

x = 7 then y = -1. So, contact point (7, -1)

2011-01-26 19:21:41 補充：
(i) (4c + 18)^2 - 20(c^2 + 6c - 2) = 0

16c^2 + 144c + 324 - 20c^2 - 120c + 40 = 0

-4c^2 + 24c + 364 = 0

c^2 - 6c - 91 = 0

(ii) y = -2x - 7

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