1. **Problem statement:** Given a complex number $z = x + yi$ satisfying $$\left|\frac{z - (4 - 6i)}{z - (5 - 5i)}\right| = 6,$$ we want to find the radius of the locus of $z$, which is a circle. 2. **Rewrite the problem:** The expression inside the modulus is a ratio of complex numbers. The modulus of a quotient is the quotient of the moduli: $$\left|\frac{z - (4 - 6i)}{z - (5 - 5i)}\right| = \frac{|z - (4 - 6i)|}{|z - (5 - 5i)|} = 6.$$ 3. **Interpretation:** This means the distance from $z$ to the point $4 - 6i$ is 6 times the distance from $z$ to the point $5 - 5i$: $$|z - (4 - 6i)| = 6 |z - (5 - 5i)|.$$ 4. **General form:** The locus of points $z$ such that the ratio of distances to two fixed points (foci) is constant is a circle (or a line if ratio = 1). Here, ratio $k=6$. 5. **Let the foci be:** $$F_1 = (4, -6), \quad F_2 = (5, -5).$$ 6. **Distance between foci:** $$d = \sqrt{(5-4)^2 + (-5 + 6)^2} = \sqrt{1^2 + 1^2} = \sqrt{2}.$$ 7. **Formula for radius of circle with ratio $k > 1$:** The circle has center $$C = \frac{k^2 F_2 - F_1}{k^2 - 1}$$ and radius $$r = \frac{k d}{|k^2 - 1|} \sqrt{(k^2 - 1) - \left(\frac{d}{r}\right)^2}$$ but a simpler approach is to use the known formula for the radius: $$r = \frac{k d}{|k^2 - 1|} \sqrt{k^2 - 1} = \frac{k d}{k^2 - 1} \sqrt{k^2 - 1} = \frac{k d}{\sqrt{k^2 - 1}}.$$ 8. **Calculate radius:** $$k = 6, \quad d = \sqrt{2}.$$ So, $$r = \frac{6 \times \sqrt{2}}{\sqrt{6^2 - 1}} = \frac{6 \sqrt{2}}{\sqrt{36 - 1}} = \frac{6 \sqrt{2}}{\sqrt{35}} = \frac{6}{35} \sqrt{70}.$$ 9. **Final answer:** The radius of the circle is $$\boxed{\frac{6}{35} \sqrt{70}}.$$