Glad I stumbled across this answer. Thanks to Spehro Pefhanys' Answer, got me thinking and calculating to a more general approach I would like to share.
simulate this circuit – Schematic created using CircuitLab
\$ m \$ denotes a scaling ratio, which in the case presented is 10:1, \$ m=10 \$
\$ Z_{out_{MAX}} = \frac{m}{m+1}\cdot \left(R_s +\frac{R_p}{4}\right)\$
\$ Z_{out_{min}} = \frac{m}{m+1} \cdot R_s\$
\$ Z_{out_{MAX}} \$ is reached When both potentiometer wipers are in the center position \$ \alpha = \beta = 0.5 \$
\$ Z_{out_{min}} \$ is reach when both potentiometers are at either extremes.
Interesting to note is that in this configuration Impedance Variance, Spread is determined solely by the Potentiometers \$ \Delta_{Z_{out}} = Z_{out_{MAX}} -Z_{out_{min}} = \frac{m}{m+1}\cdot \frac{R_p}{4}\$
If you consider \$ Z_{load} >>> Z_{out} \$ then: \$ \frac{V_{out}}{V_{in}}=\frac{m}{m+1}\cdot\left(\alpha +\frac{\beta}{m}\right)\$
\$ Z_{in} \approx R_{p1} // R_{p2} \$ , where Rp1 and Rp2 are the potentiometers represented at Rp and m·Rp in the diagram.
The circuit input impedance is relatively constant, only slightly altered with different wiper positions or even different loads.
Minor \$ \Delta_{Z_{out}} \$, Impedance variance improvements:
As can be demonstrated the fine/coarse ratio is defined by \$ m , R_{s2} = m \cdot R_{s1} \$, The impedance swing is defined solely by the potentiometers \$ R_{p2} = m \cdot R_{p1} \$
The formulas presented scale the potentiometers with the ratio \$ m \$ although they need not be. As presented initially by Spehro, they can be of the same "value". Not scaling the values increases input load but can slightly improve the impedance variance. By how much can be approximated as follows.
Let \$ f(x) = \Delta_{Z_{out}} = \frac{x}{x+1}\cdot \frac{R_p}{4}\$
\$ f'(x) = \frac{R_p}{4\left(1+x\right)^2} \$
evaluate both \$ f(m) \$ and \$ f'(m) \$ we can define a linear function :
\$ g(k) = k\cdot f'(m) + b \$
where b is found by solving \$ g(m) = f(m) \$. Now we will have a linear function \$ g(k)\$ which will approximate the impedance variance given a factor k between the potentiometers \$ R_{p2} = k\cdot R_{p1}\$ while maintaining the factor \$ m \$ for the coarse/fine ratio.
For the example provided by Spehro, \$ m=10, R_p = 0.5 k\Omega \$
\$ g(k) = \frac{k}{968} + \frac{100}{968} \$
the improvement from using two \$ 500 \Omega \$ pots, \$ g(1) \approx 104 \Omega \$ instead of a \$ 500 \Omega \$ and \$ 5k \Omega \$ pot, \$ g(10) \approx 114 \Omega \$ is an impedance swing improvement of \$ \approx 10\Omega \$
Actually, if you are willing to have an input impedance of approximately \$ \approx 250\Omega \$ you can achieve a tighter impedance swing by using \$ 250 \Omega \$ and \$ 2k5 \Omega \$ pots which would reduce the impedance variation down to \$ \Delta_{Z_{out}} \approx 57\Omega \$
Some formulas for same layout but with resistors and pots that are not bound by a ratio
The output impedance can be calculated as follows:
\$ Z_{out} = \left(R_{p1}+R_{s1}\right) //\left(R_{p2}+R_{s2}\right)= \frac{\left(R_{p1}+R_{s1}\right)\cdot\left(R_{p2}+R_{s2}\right)}{R_{p1}+R_{s1}+R_{p2}+R_{s2}}\$
Where : \$ R_{p1} = R_{p1_{Total}}\cdot(1-\alpha)\alpha\$
, being \$ \alpha \$ the wiper position \$ \{0..1\} \$
\$ Z_{out_{MAX}} = \frac{\left(R_{p1T}+4R_{s1}\right)\left(R_{p2T}+4R_{s2}\right)}{4\left(R_{p1T}+4R_{s1}+R_{p2T}+4R_{s2}\right)}\$ When both potentiometer wipers are in the center position \$ \alpha = 0.5 \$
\$ Z_{out_{min}} = \frac{R_{s1}R_{s2}}{R_{s1}+R_{s2}} \$
If you consider \$ Z_{load} >>> Z_{out} \$ then: \$ \frac{V_{out}}{V_{in}}=\frac{\alpha R_{s2}+\beta R_{s1}}{R_{s1}+R_{s2}}\$
Just thought I may share my exploration and generalization of the answer.
