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The equations of the two circles are $x^2+y^2=36$ and $(x-5)^2 +y^2=16$. The problem asks to find a common tangent line in point-slope form.

I've tried drawing a diagram and finding the distance between the points of tangency, but that did not help in finding a point of tangency or the slope of the lines.

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  • $\begingroup$ Draw the common tangent & think about where it intersects the $x$ axis. Now draw the normals (from the centers of the circles) to their respective tangent points. You should have two similar right angled triangles ? now you can set up a ratio ? ... more hints on request $\endgroup$
    – Donald Splutterwit
    Mar 22, 2017 at 0:15

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A hint (after Donald Splutterwit):

enter image description here

(Sorry... one cannot post graphics in a comment...)

So solve for $x$:

${4 \over x} = {6 \over 5+x}$

to find $x = 10$.

Now what?....

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The two tangents either side of the circles meet at a point, and the line through the circle centres bisects that meeting angle. The distance to that point is in ratio to the circle diameters, so given that the circle centres given here are $5$ units apart, and are radius $6$ and $4$, the meeting point of the two tangents will be $10$ units beyond the $4$ radius circle, giving a point on both tangents of $(15,0)$.

Then the slope can be calculated from a right triangle to the touch point on the near triangle as $\pm\frac 4a $ given $a^2 = 10^2-4^2$.

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