If the circle $$x^2 + y^2 - 2gx + 6y - 19c = 0$$, $$g, c \in \mathbb{R}$$ passes through the point $$(6, 1)$$ and its centre lies on the line $$x - 2cy = 8$$, then the length of intercept made by the circle on $$x$$-axis is

The curve is $$y = x^3 + 3x^2 + 5$$, so $$\dfrac{dy}{dx} = 3x^2 + 6x$$.

The tangent at $$(x_1, y_1)$$ is: $$y - y_1 = (3x_1^2 + 6x_1)(x - x_1)$$.

Since it passes through the origin $$(0, 0)$$: $$-y_1 = (3x_1^2 + 6x_1)(0 - x_1) = -x_1(3x_1^2 + 6x_1)$$.

So $$y_1 = 3x_1^3 + 6x_1^2$$ $$-(1)$$

Since $$(x_1, y_1)$$ lies on the curve: $$y_1 = x_1^3 + 3x_1^2 + 5$$ $$-(2)$$

From $$(1)$$ and $$(2)$$: $$x_1^3 + 3x_1^2 + 5 = 3x_1^3 + 6x_1^2$$

$$2x_1^3 + 3x_1^2 - 5 = 0$$

We can factor this as $$(x_1 - 1)(2x_1^2 + 5x_1 + 5) = 0$$.

The quadratic $$2x_1^2 + 5x_1 + 5 = 0$$ has discriminant $$25 - 40 = -15 < 0$$, so the only real solution is $$x_1 = 1$$.

$$y_1 = 1 + 3 + 5 = 9$$. So the point is $$(1, 9)$$.

Now we check each option:

Option A: $$x^2 + \dfrac{y^2}{81} = 1 + \dfrac{81}{81} = 1 + 1 = 2$$ ✓ The point lies on this curve.

Option B: $$\dfrac{y^2}{9} - x^2 = \dfrac{81}{9} - 1 = 9 - 1 = 8$$ ✓ The point lies on this curve.

Option C: $$y = 4x^2 + 5 = 4(1) + 5 = 9$$ ✓ The point lies on this curve.

Option D: $$\dfrac{x}{3} - y^2 = \dfrac{1}{3} - 81 = -\dfrac{242}{3} \neq 0$$ ✗ The point does NOT lie on this curve.

The answer is Option D: $$\dfrac{x}{3} - y^2 = 0$$.