Geometry/Trigonometry
posted by Sam .
Circles with centers (2,1) and (8,9) have radii 1 and 9, respectively. The equation of a common external tangent to the circles can be written in the form y=mx+b with m < 0. What is b?
I drew the diagram and the tangents but to no avail. I cannot seem to find the points where the tangent hits both circles because if I do, the problem would be solved. Help is appreciated, thanks in advance

Both circles are tangent to the xaxis, at (2,0) and (8,0).
The line joining their centers is inclined at an angle θ such that
tanθ = (91)/(82) = 4/3
So, the two tangent lines meet at an angle 2θ. Thus, our slope m of the other line is
tan2θ = (8/3)/(116/9) = 24/7
Now we have a line
y = 24/7 x + b
which must be tangent to both circles. That is, if we look for where the line intersects the circle, there must be a single solution.
Taking the first circle, we need
(x2)^2 + (y1)^2 = 1
(x2)^2 + (24/7 x+b1)^2 = 1
x^24x+4 + 576/49 x^2  48/7 (b1)x + (b1)^2 = 1
625/49 x^2  (48b20)/7 x + (b^22b+4) = 0
For that to have a single solution, the discriminant must be zero, so
((48b20)/7)^2  4(625/49)(b^22b+4) = 0
b = 30/7 and 80/7
we want the smaller value. The larger one will be on the other side of the circle. So,
y = 24/7 x + 30/7
To see the graphs, visit
http://www.wolframalpha.com/input/?i=plot+%28x2%29^2%2B%28y1%29^2%3D1+and+%28x8%29^2%2B%28y9%29^2%3D81%2C+y%3D24%2F7+x+%2B+30%2F7 
Wow thanks a lot that helped!
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