急! F5 Equations of Circles 8q3

16a) Sub y = mx + 2 into C:

(x - 6)2 + (mx - 2)2 = 8

x2 - 12x + 36 + m2x2 - 4mx + 4 = 8

(m2 + 1)x2 - (4m + 12)x + 32 = 0

Since L is tangent to C, this quad. equation has its discriminant = 0:

(4m + 12)2 - 4(32)(m2 + 1) = 0

(m + 3)2 - 8(m2 + 1) = 0

m2 + 6m + 9 - 8m2 - 8 = 0

-7m2 + 6m + 1 = 0

7m2 - 6m - 1 = 0

b) From (a), m = -1/7 or 1

So L can be y = x + 2 or y = -x/7 + 2, i.e. x + 7y - 14 = 0

17a) y = mx - 2

b) Sub y = mx - 2 into C:

x2 + (mx - 2)2 - 6(mx - 2) - 7 = 0

x2 + m2x2 - 4mx + 4 - 6mx + 12 - 7 = 0

(m2 + 1)x2 - 10mx + 9 = 0

Since L is tangent to C, discriminant of this quad eqn is 0:

(-10)2 - 4(9)(m2 + 1) = 0

100 - 36(m2 + 1) = 0

m2 = 16/9

m = 4/3 or -4/3

c) First: y = 4x/3 - 2, i.e. 4x - 3y - 6 = 0

Second: y = -4x/3 - 2, i.e. 4x + 3y + 6 = 0