Let the tangent to the circle $$x^2 + y^2 = 25$$ at the point $$R(3, 4)$$ meet $$x$$-axis and $$y$$-axis at point $$P$$ and $$Q$$, respectively. If $$r$$ is the radius of the circle passing through the origin $$O$$ and having centre at the incentre of the triangle $$OPQ$$, then $$r^2$$ is equal to:

The tangent to the circle $$x^2 + y^2 = 25$$ at the point $$R(3, 4)$$ has the equation $$3x + 4y = 25$$.

To find where this meets the $$x$$-axis, set $$y = 0$$: $$3x = 25$$, so $$P = \left(\frac{25}{3}, 0\right)$$.

To find where it meets the $$y$$-axis, set $$x = 0$$: $$4y = 25$$, so $$Q = \left(0, \frac{25}{4}\right)$$.

Now we have the triangle $$OPQ$$ with vertices $$O = (0, 0)$$, $$P = \left(\frac{25}{3}, 0\right)$$, $$Q = \left(0, \frac{25}{4}\right)$$.

The side lengths are: $$OP = \frac{25}{3}$$, $$OQ = \frac{25}{4}$$, and $$PQ = \sqrt{\left(\frac{25}{3}\right)^2 + \left(\frac{25}{4}\right)^2} = 25\sqrt{\frac{1}{9} + \frac{1}{16}} = 25\sqrt{\frac{25}{144}} = \frac{25 \times 5}{12} = \frac{125}{12}$$.

For the incentre formula, we label: side $$a = PQ = \frac{125}{12}$$ (opposite vertex $$O$$), side $$b = OQ = \frac{25}{4}$$ (opposite vertex $$P$$), side $$c = OP = \frac{25}{3}$$ (opposite vertex $$Q$$).

The incentre is $$I = \frac{a \cdot O + b \cdot P + c \cdot Q}{a + b + c}$$.

First, $$a + b + c = \frac{125}{12} + \frac{25}{4} + \frac{25}{3} = \frac{125}{12} + \frac{75}{12} + \frac{100}{12} = \frac{300}{12} = 25$$.

$$I_x = \frac{a \cdot 0 + b \cdot \frac{25}{3} + c \cdot 0}{25} = \frac{\frac{25}{4} \cdot \frac{25}{3}}{25} = \frac{625}{300} = \frac{25}{12}$$.

$$I_y = \frac{a \cdot 0 + b \cdot 0 + c \cdot \frac{25}{4}}{25} = \frac{\frac{25}{3} \cdot \frac{25}{4}}{25} = \frac{625}{300} = \frac{25}{12}$$.

So the incentre is $$I = \left(\frac{25}{12}, \frac{25}{12}\right)$$.

The circle passes through the origin $$O(0,0)$$ and has its centre at $$I$$. So the radius $$r$$ equals the distance from $$I$$ to $$O$$.

$$r^2 = \left(\frac{25}{12}\right)^2 + \left(\frac{25}{12}\right)^2 = 2 \cdot \frac{625}{144} = \frac{1250}{144} = \frac{625}{72}$$.

The answer is $$\frac{625}{72}$$, which is Option C.