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Let the contact point have positive horizontal coordinate $x_0$. Equality of slopes between the tangent line and the logarithmic curve requires $k=1/x_0$. The point must also satisfy both equations, giving $\ln x_0=k{x}_0+1=2$. HenceThe contact point is $(e^2,2)$, and the resulting line $y=e^{-2}x+1$ has exactly the required derivative there.

Use completing the square in each coordinate:Comparing with the standard circle equation gives

For the parametric curve, $dx/dt=e^t\ne 0$ and $dy/dt=\cos t$. Therefore $dy/dx=e^{-t}\cos t$. The condition $x=e^t=1$ gives $t=0$, so