No, they are the gradients at each point on the respective circles.

There's very likely some circle theorems that can be used to solve this problem easily (the two circles are tangent at (1, 0) and you know the radii of each circle) but in the meantime your stuck with this approach:

Let the gradient of the tangent be m.

Let the point on the circle be (a, b). Then:

..... (1)

.... (2)

Let the point on the circle be (e, f). Then:

..... (3)

.... (4)

Also, .... (5)

Solve equations (1) to (5) simultaneously for a, b, e, f. I'd suggest eliminating m and solve the resulting four equations in the unknowns a,,b, e, f.