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 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
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Wow thanks a lot that helped!