Let the circle touch the line x-y+1=0, have the centre on the positive x -axis, and cut off a chord of length $$\frac{4}{\sqrt{13}}$$ along the line -3x+2y=1. Let H be the hyperbola $$\frac{x^{2}}{\alpha^{2}}-\frac{y^{2}}{\beta^{2}}=1$$ , whose one of the foci is the centre of C and the length of the transverse axis is the diameter of C. Then $$2\alpha^{2}+3\beta^{2}$$ is equal to ______

The circle touches the line $$x - y + 1 = 0$$ and has its center on the positive x-axis. Let the center be $$(h, 0)$$ with $$h > 0$$. The distance from the center to the line $$x - y + 1 = 0$$ equals the radius $$r$$. The distance formula gives:

$$\frac{|h - 0 + 1|}{\sqrt{1^2 + (-1)^2}} = \frac{|h + 1|}{\sqrt{2}} = r \quad \text{(1)}$$

Since $$h > 0$$, $$|h + 1| = h + 1$$, so $$r = \frac{h + 1}{\sqrt{2}}$$.

The circle cuts a chord of length $$\frac{4}{\sqrt{13}}$$ along the line $$-3x + 2y - 1 = 0$$. The perpendicular distance $$d$$ from the center $$(h, 0)$$ to this line is:

$$d = \frac{|-3h + 2 \cdot 0 - 1|}{\sqrt{(-3)^2 + 2^2}} = \frac{|-3h - 1|}{\sqrt{13}} = \frac{|3h + 1|}{\sqrt{13}} \quad \text{(2)}$$

Since $$h > 0$$, $$|3h + 1| = 3h + 1$$, so $$d = \frac{3h + 1}{\sqrt{13}}$$.

For a chord of length $$l$$, the relationship between $$d$$, $$r$$, and $$l$$ is:

$$\left( \frac{l}{2} \right)^2 + d^2 = r^2$$

Given $$l = \frac{4}{\sqrt{13}}$$, so $$\frac{l}{2} = \frac{2}{\sqrt{13}}$$. Substituting:

$$\left( \frac{2}{\sqrt{13}} \right)^2 + d^2 = r^2 \implies \frac{4}{13} + d^2 = r^2 \quad \text{(3)}$$

Substituting $$d$$ from (2):

$$\frac{4}{13} + \left( \frac{3h + 1}{\sqrt{13}} \right)^2 = r^2 \implies \frac{4}{13} + \frac{(3h + 1)^2}{13} = r^2 \implies r^2 = \frac{4 + (3h + 1)^2}{13} \quad \text{(4)}$$

From (1), $$r^2 = \left( \frac{h + 1}{\sqrt{2}} \right)^2 = \frac{(h + 1)^2}{2}$$. Equating to (4):

$$\frac{(h + 1)^2}{2} = \frac{4 + (3h + 1)^2}{13}$$

Cross-multiplying by 26:

$$13(h + 1)^2 = 2 \left[ 4 + (3h + 1)^2 \right]$$

Expanding:

$$13(h^2 + 2h + 1) = 2[4 + 9h^2 + 6h + 1] \implies 13h^2 + 26h + 13 = 2(9h^2 + 6h + 5) \implies 13h^2 + 26h + 13 = 18h^2 + 12h + 10$$

Rearranging:

$$13h^2 + 26h + 13 - 18h^2 - 12h - 10 = 0 \implies -5h^2 + 14h + 3 = 0$$

Multiplying by -1:

$$5h^2 - 14h - 3 = 0$$

Solving the quadratic equation:

$$h = \frac{14 \pm \sqrt{(-14)^2 - 4 \cdot 5 \cdot (-3)}}{10} = \frac{14 \pm \sqrt{196 + 60}}{10} = \frac{14 \pm \sqrt{256}}{10} = \frac{14 \pm 16}{10}$$

So $$h = 3$$ or $$h = -0.2$$. Since $$h > 0$$, $$h = 3$$.

Radius $$r = \frac{3 + 1}{\sqrt{2}} = \frac{4}{\sqrt{2}} = 2\sqrt{2}$$. The diameter is $$2r = 4\sqrt{2}$$.

The hyperbola is $$\frac{x^2}{\alpha^2} - \frac{y^2}{\beta^2} = 1$$. One focus is the center of the circle, $$(3, 0)$$. For a hyperbola, foci are at $$(\pm c, 0)$$ where $$c = \sqrt{\alpha^2 + \beta^2}$$. Thus:

$$\sqrt{\alpha^2 + \beta^2} = 3 \quad \text{(5)}$$

The transverse axis length is $$2\alpha$$, equal to the diameter $$4\sqrt{2}$$:

$$2\alpha = 4\sqrt{2} \implies \alpha = 2\sqrt{2} \quad \text{(6)}$$

Substituting $$\alpha = 2\sqrt{2}$$ into (5):

$$\sqrt{(2\sqrt{2})^2 + \beta^2} = 3 \implies \sqrt{8 + \beta^2} = 3 \implies 8 + \beta^2 = 9 \implies \beta^2 = 1$$

Now compute $$2\alpha^2 + 3\beta^2$$:

$$\alpha^2 = (2\sqrt{2})^2 = 8, \quad \beta^2 = 1$$

$$2 \times 8 + 3 \times 1 = 16 + 3 = 19$$

The answer is 19.