If one of the diameters of the circle $$x^2 + y^2 - 10x + 4y + 13 = 0$$ is a chord of another circle $$C$$, whose center is the point of intersection of the lines $$2x + 3y = 12$$ and $$3x - 2y = 5$$, then the radius of the circle $$C$$ is

We begin by finding the centre and radius of the circle whose equation is $$x^2 + y^2 - 10x + 4y + 13 = 0$$. Completing the square yields $$(x^2 - 10x + 25) + (y^2 + 4y + 4) + 13 - 25 - 4 = 0$$, which simplifies to $$(x - 5)^2 + (y + 2)^2 = 16$$. Hence, the centre is $$(5, -2)$$ and the radius is $$r = 4$$, making the length of its diameter $$8$$ and the half-length of the diameter (the chord length) equal to $$4$$.

The centre of the circle $$C$$ is determined by the intersection of the lines $$2x + 3y = 12$$ and $$3x - 2y = 5$$. Solving these simultaneously, multiplying the first by 2 gives $$4x + 6y = 24$$ and multiplying the second by 3 gives $$9x - 6y = 15$$. Adding these equations yields $$13x = 39$$, so $$x = 3$$. Substituting back into $$2x + 3y = 12$$ gives $$2(3) + 3y = 12$$, hence $$y = 2$$. Therefore, the centre of $$C$$ is $$(3, 2)$$.

The given diameter of the first circle serves as a chord of circle $$C$$ and its midpoint is the centre $$(5, -2)$$ of the first circle. The perpendicular distance $$d$$ from the centre of $$C$$ to this chord is found by the distance formula: $$d = \sqrt{(5 - 3)^2 + (-2 - 2)^2} = \sqrt{20} = 2\sqrt{5}\,. $$

In any circle, the relationship between the radius $$R$$, the half-length of a chord $$l$$, and the perpendicular distance $$d$$ from the centre to the chord is given by $$R^2 = d^2 + l^2$$. Substituting $$d^2 = 20$$ and $$l = 4$$ gives $$R^2 = 20 + 16 = 36$$, so $$R = 6$$.

Therefore, the radius of circle $$C$$ is 6.