數學勁既人入黎.

Center and radius of the 2 circles are (3, 0) , 5 and (-3, 0), 7 respectively.
Let the 2 common tangents intersect at M( a, 0).
By similar triangle, (-3 - a)/5 = (3 - a)/7
-21 - 7a = 15 - 5a
-2a = 36
a = - 18. That is M( - 18, 0) is on the tangent....................(1)
Let angle between the tangent and the x - axis be +/- z,
so sin z = 5/15 = 1/3, that is tan z = 1/sqrt 8 = slope of tangent..................(2).
From (1) and (2), equations of the 2 common tangents are:
y = (x + 18)/sqrt 8 and y = -( x + 18)/sqrt 8, these can be re-arranged as :
2 sqrt 2 y = x + 18 and
- 2 sqrt 2 y = x + 18.
There are 4 points of contact, 2 for each circle.
Take tangent 2 sqrt 2 y = x + 18, let the point of contact with the 1st circle be (m, n).
so 2 sqrt 2 n = m + 18 ..................(3)
slope of tangent x slope of point of contact and center of circle = -1
that is (1/2 sqrt 2)[(n - 0)/(m - 3)] = -1
n = 2 sqrt 2 ( 3 - m)....................(4)
Solving (3) and (4) we get
8(3 - m) = m + 18
24 - 8m = m + 18
6 = 9m
m = 2/3 , n = 2 sqrt 2(3 - 2/3) = (14 sqrt 2)/3
so one of the points of contact is [ 2/3, (14 sqrt 2)/3].
By symmetry about the x -axis, the other point of contact is
[2/3, (- 14 sqrt 2)/3].
You can apply the same method to the other circle to find the remaining 2 points of contact.

C1 : x^2 + y^2 – 6x – 16 = 0
C2 : x^2 + y^2 + 6x – 40 = 0
Let the tangent be y = mx + c … (1)
Sub (1) into C1 gives
(1 + m^2)x^2 + (2mc – 6)x + (c^2 – 16) = 0 … (a)
For tangent, the quadratic equation must have repeated root, therefore
(2mc – 6)^2 – 4(1 + m^2)(c^2 – 16) = 0
Or 16m^2 – c^2 – 6mc + 25 = 0 … (2)
Sub (1) into C2 gives
(1 + m^2)x^2 + (2mc + 6)x + (c^2 – 40) = 0 … (b)
For tangent, the quadratic equation must have repeated root, therefore
(2mc + 6)^2 – 4(1 + m^2)(c^2 – 40) = 0
Or 40m^2 – c^2 + 6mc + 49 = 0 … (3)
(3) – (2) yields
24m^2 + 12mc + 24 = 0 or
c = –2(m^2 + 1)/m …(4)
(3) + (2) yields
56 m^2 – 2 c^2 + 74 = 0 … (5)
Put (4) into (5),
48m^4 + 58m^2 – 8 = 0
2(3m^2 + 4)(8m^2 – 1) = 0
For real m, m = +/– 1/sqrt(8)

When m = 1/ sqrt(8), c= –18/sqrt(8)
Tangent 1 is y = 1/sqrt(8) x – 18 / sqrt(8) or 2sqrt(2)y – x + 18 =0
Sub into (a) gives 9x^2 – 84x + 196 = 0
X= 14/3 and (1) gives y = – 10 sqrt(2) / 3
Sub into (b) gives 9x^2 + 12x + 4 = 0
X= 2/3 and (1) gives y = – 14 sqrt(2) / 3

When m = –1/ sqrt(8), c= 18/sqrt(8)
Tangent 2 is y = –1/sqrt(8) x + 18 / sqrt(8) or 2sqrt(2)y + x – 18 =0
Sub into (a) gives 9x^2 – 84x + 196 = 0
X= 14/3 and (1) gives y = 10 sqrt(2) / 3
Sub into (b) gives 9x^2 + 12x + 4 = 0
X= 2/3 and (1) gives y = 14 sqrt(2) / 3