Let a circle passing through $$(2, 0)$$ have its centre at the point $$(h, k)$$. Let $$(x_c, y_c)$$ be the point of intersection of the lines $$3x + 5y = 1$$ and $$(2 + c)x + 5c^2y = 1$$. If $$h = \lim_{c \to 1} x_c$$ and $$k = \lim_{c \to 1} y_c$$, then the equation of the circle is :

Find intersection of $$3x + 5y = 1$$ and $$(2+c)x + 5c^2y = 1$$.

At $$c = 1$$: both become $$3x + 5y = 1$$, so we need limits.

Subtracting: $$(2+c)x + 5c^2y - 3x - 5y = 0 \implies (c-1)x + 5(c^2-1)y = 0$$.

$$(c-1)x + 5(c-1)(c+1)y = 0$$. For $$c \neq 1$$: $$x + 5(c+1)y = 0$$.

As $$c \to 1$$: $$x + 10y = 0 \implies x = -10y$$.

Substituting into $$3x + 5y = 1$$: $$-30y + 5y = 1 \implies -25y = 1 \implies y = -\frac{1}{25}$$.

$$x = -10 \times (-\frac{1}{25}) = \frac{2}{5}$$.

So $$h = \frac{2}{5}$$, $$k = -\frac{1}{25}$$.

Radius = distance from centre $$(h,k)$$ to $$(2,0)$$:

$$r^2 = (2-\frac{2}{5})^2 + (0+\frac{1}{25})^2 = (\frac{8}{5})^2 + (\frac{1}{25})^2 = \frac{64}{25} + \frac{1}{625} = \frac{1600+1}{625} = \frac{1601}{625}$$

Circle equation: $$(x-\frac{2}{5})^2 + (y+\frac{1}{25})^2 = \frac{1601}{625}$$

Expanding: $$x^2 - \frac{4x}{5} + \frac{4}{25} + y^2 + \frac{2y}{25} + \frac{1}{625} = \frac{1601}{625}$$

$$x^2 + y^2 - \frac{4x}{5} + \frac{2y}{25} = \frac{1601}{625} - \frac{4}{25} - \frac{1}{625} = \frac{1601 - 100 - 1}{625} = \frac{1500}{625} = \frac{12}{5}$$

Multiplying by 25: $$25x^2 + 25y^2 - 20x + 2y = 60$$

$$25x^2 + 25y^2 - 20x + 2y - 60 = 0$$.

The correct answer is Option 4.