A little bit of analytic geometry

The equations of the circle and the ellipse are

$\begin{cases} x^2+y^2 = 36 & ...(1) \\ \dfrac {x^2}{25} + \dfrac {(y-y_0)^2}4 = 1 & ...(2) \end{cases}$

At the tangent point $P(x_1,y_1)$ the gradients of the ellipse and the circle are the same. From differentiation with respect to $x$ , we have:

$\begin{cases} (1): \quad 2x + 2y \cdot \dfrac {dy}{dx} = 0 & \implies \dfrac {dy}{dx} = - \dfrac xy \\ (2): \quad \dfrac {2x}{25} + \dfrac {2(y-y_0)}4 \cdot \dfrac {dy}{dx} = 0 & \implies \dfrac {dy}{dx} = - \dfrac {4x}{25(y-y_0)} \end{cases}$

$\implies \frac {x_1}{y_1} = \frac {4x_1}{25(y_1-y_0)} \implies y_1 = \frac {25}{21}y_0$

From $(1) - 25 \times (2)$ :

$\begin{aligned} y_1^2 - \frac {25}4 (y_1 - y_0)^2 & = 36 - 25 \\ \frac {25^2}{21^2}y_0^2 - \frac {25}4 \left(\frac 4{21} y_0 \right)^2 & = 11 \\ \frac {25}{21} y_0^2 & = 11 \\ \implies y_0 & = - \frac {\sqrt{231}}5 & \small \blue{\text{Since }y_0 < 0} \\ y_1 & = \frac {25}{21}y_0 = - 5 \sqrt{\frac {11}{21}} \\ (1): \quad \quad x_1 & = \sqrt{36-y_1^2} = \sqrt{\frac {481}{21}} \\ \implies x_1 - y_1 - y_0 & = \sqrt{\frac {481}{21}} + 5 \sqrt{\frac {11}{21}} + \frac {\sqrt{231}}5 \\ & \approx \boxed{11.4} \end{aligned}$

Let the ellipse have the equation $\large \frac{x^2}{25} + \frac{(y-y_{0})^2}{4} = 1$ and the circle be $x^2 + y^2 = 36.$ If we substitute $x^2 = 36-y^2$ into the elliptical equation, we now obtain a quadratic in $y:$

$\large \frac{36-y^2}{25} + \frac{(y-y_{0})^2}{4} = 1;$

or $4(36-y^2) + 25(y-y_{0})^2 = 100;$

or $21y^2 - 50y_{0}y +(25y^{2}_{0}+44)=0;$

or $\large y = \frac{50y_{0} \pm \sqrt{2500y^{2}_{0} - 4(21)(25y^{2}_{0}+44)}}{42} = \frac{50y_{0} \pm \sqrt{400y^{2}_{0}-3696}}{42}.$

If the ellipse is to intersect the circle in only one value for $y$ , then the discriminant above shall equal zero $\Rightarrow y^{2}_{0} = \frac{3696}{400} \Rightarrow y_{0} = \pm 3.04$ (of which we only admit the negative root as the ellipse lies entirely below the $x-$ axis). This in turn yields the right-tangency point $y_{1} = \frac{50(-3.04)}{42} = -3.619, x_{1} = \sqrt{36-3.619^2} = 4.786.$ The final desired result computes to:

$x_{1}-y_{1}-y_{0} = 4.786 -(-3.619) - (-3.04) = \boxed{11.445}.$