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 $$10\alpha^2$$ where the point $$(1,\alpha)$$ lies on the largest circle centered at $$(2,0)$$ inscribed in the ellipse $$\frac{x^2}{36} + \frac{y^2}{16} = 1$$. Since the circle is centered at $$(2,0)$$ with radius $$r$$, its equation is $$(x-2)^2 + y^2 = r^2$$, and it must be tangent internally to the ellipse so that at the point of tangency the ellipse and the circle share the same tangent line.

Next, from the ellipse we have $$y^2 = 16\left(1 - \frac{x^2}{36}\right) = 16 - \frac{4x^2}{9}$$, and substituting this into the circle equation gives

$$ (x-2)^2 + 16 - \frac{4x^2}{9} = r^2 $$

This expands to $$x^2 - 4x + 4 + 16 - \frac{4x^2}{9} = r^2$$ and simplifies to $$\frac{5x^2}{9} - 4x + 20 = r^2$$, which can be written as $$5x^2 - 36x + 9(20 - r^2) = 0$$.

Requiring tangency means the discriminant of this quadratic in $$x$$ is zero, namely $$36^2 - 4 \cdot 5 \cdot 9(20 - r^2) = 0$$. Therefore,

$$ 1296 - 180(20 - r^2) = 0\,,\quad1296 - 3600 + 180r^2 = 0\,,\quad180r^2 = 2304\,,\quad r^2 = \frac{2304}{180} = \frac{64}{5}\,. $$

Since the point $$(1,\alpha)$$ lies on this circle, substituting into $$(x-2)^2 + y^2 = r^2$$ gives $$(1-2)^2 + \alpha^2 = r^2$$, which yields $$1 + \alpha^2 = \frac{64}{5}$$ and hence $$\alpha^2 = \frac{64}{5} - 1 = \frac{59}{5}\,.$$

Therefore, $$10\alpha^2 = 10 \times \frac{59}{5} = 2 \times 59 = 118\,. $$

The correct answer is $$\mathbf{118}$$.