一條關於Circles的簡單A.Maths.問題(2)

http://pstock.idv.tw/files-pstock/9042201.pdf

Since the centre of the 2 circles are on the x -axis, by property of tangent, the common tangents must intersects on the x -axis, let the intersecting point be (h, 0).
Radius of the 2 circles are 7 and 5, by similar triangle,
(h -3)/5 = (h + 3)/7, so h = 18, therefore, equation of the common tangents will be:
y = m(x - 18). Sub into equation of the first circle, we get
x^2 + (mx - 18m)^2 -6x -16 = 0
(1 + m^2)x^2 - (36m^2 + 6)x + (324m^2 - 16) = 0.............(1)
For tangent, delta = 0, so
(36m^2 + 6)^2 - 4(1 + m^2)(324m^2 - 16) = 0
(18m^2 + 3)^2 - (1 + m^2)(324m^2 - 16) = 0
324m^4 + 9 + 108m^2 - 324m^2 + 16 - 324m^4 + 16m^2 = 0
200m^2 - 25 = 0
m^2 = 25/200 = 1/8, m = +/- sqrt(1/8).
so the common tangents are:
+/1(2sqrt2)y = x - 18
Since delta = 0, the x-coordinate of the points of contact of circle C1 will be the root of equation (1) above. That is (36 /8 + 6)/2(1 + 1/8)
= 84/18 = 14/3. From equation of the circle, you can find the y-coordinate to be +/- (10 sqrt2)/3.
Using point of division 6 : 15, you can get the point of contacts of the other circle.
14/3 = [(15x + (6)(18)]/(6 + 15) = (15x + 108)/21
98 = 15x + 108
x = -10/15 = -2/3, similarly, you can find the y-coordinate from equation of C2.