Let $$C$$ be the largest circle centred at $$(2, 0)$$ and inscribed in the ellipse $$\frac{x^2}{36} + \frac{y^2}{16} = 1$$. If $$(1, \alpha)$$ lies on $$C$$, then $$10\alpha^2$$ is equal to ______

We need to find the largest circle centered at $$(2, 0)$$ inscribed in the ellipse $$\frac{x^2}{36} + \frac{y^2}{16} = 1$$, and then compute $$10\alpha^2$$ given that $$(1, \alpha)$$ lies on this circle.

The equation of the circle is $$(x-2)^2 + y^2 = r^2$$, and for it to be inscribed in the ellipse it must be tangent to the ellipse.

From the ellipse we have $$y^2 = 16\left(1 - \frac{x^2}{36}\right) = \frac{16(36 - x^2)}{36} = \frac{4(36 - x^2)}{9}\,.$$

Substituting this into the circle’s equation gives $$(x-2)^2 + \frac{4(36 - x^2)}{9} = r^2$$ which can be rewritten as $$9(x-2)^2 + 4(36 - x^2) = 9r^2,$$ $$9x^2 - 36x + 36 + 144 - 4x^2 = 9r^2,$$ $$5x^2 - 36x + 180 = 9r^2\,.$$

For the circle to be tangent to the ellipse, the resulting quadratic in $$x$$ must have a double root, so its discriminant is zero: $$5x^2 - 36x + (180 - 9r^2) = 0,$$ $$\Delta = 36^2 - 4(5)(180 - 9r^2) = 0,$$ $$1296 - 20(180 - 9r^2) = 0,$$ $$1296 - 3600 + 180r^2 = 0,$$ $$180r^2 = 2304,$$ $$r^2 = \frac{2304}{180} = \frac{64}{5}\,.$$

Finally, since $$(1, \alpha)$$ lies on the circle $$(x-2)^2 + y^2 = \frac{64}{5}$$, we substitute $$x = 1$$ and $$y = \alpha$$ to get $$(1 - 2)^2 + \alpha^2 = \frac{64}{5},$$ $$1 + \alpha^2 = \frac{64}{5},$$ $$\alpha^2 = \frac{64}{5} - 1 = \frac{59}{5},$$ $$10\alpha^2 = 10 \times \frac{59}{5} = 118\,.$$

The answer is 118.