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?

Question

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

Answer 1

Both circles are tangent to the x-axis, at (2,0) and (8,0).

The line joining their centers is inclined at an angle θ such that

tanθ = (9-1)/(8-2) = 4/3

So, the two tangent lines meet at an angle 2θ. Thus, our slope m of the other line is
tan2θ = (8/3)/(1-16/9) = -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

(x-2)^2 + (y-1)^2 = 1
(x-2)^2 + (-24/7 x+b-1)^2 = 1
x^2-4x+4 + 576/49 x^2 - 48/7 (b-1)x + (b-1)^2 = 1
625/49 x^2 - (48b-20)/7 x + (b^2-2b+4) = 0

For that to have a single solution, the discriminant must be zero, so

((48b-20)/7)^2 - 4(625/49)(b^2-2b+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+%28x-2%29^2%2B%28y-1%29^2%3D1+and+%28x-8%29^2%2B%28y-9%29^2%3D81%2C+y%3D-24%2F7+x+%2B+30%2F7

Answer 2

Wow thanks a lot that helped!