1. **State the problem:** We have a circle $C$ with center on the positive $x$-axis, tangent to the line $x - y + 1 = 0$, and cutting a chord of length $\frac{4}{\sqrt{13}}$ on the line $-3x + 2y = 1$. A hyperbola $H$ is defined by $\frac{x^2}{\alpha^2} - \frac{y^2}{\beta^2} = 1$, with one focus at the center of $C$ and transverse axis length equal to the diameter of $C$. We need to find $2\alpha^2 + 3\beta^2$. 2. **Find the center and radius of circle $C$:** Let the center be $O = (a,0)$ with $a > 0$. - The circle is tangent to the line $x - y + 1 = 0$. The distance from center to this line equals the radius $r$: $$r = \frac{|a - 0 + 1|}{\sqrt{1^2 + (-1)^2}} = \frac{|a + 1|}{\sqrt{2}} = \frac{a + 1}{\sqrt{2}}$$ (since $a > 0$, $a+1 > 0$). 3. **Use the chord length condition:** The chord is cut on the line $-3x + 2y = 1$. Rewrite as $-3x + 2y - 1 = 0$. - Distance from center to this line: $$d = \frac{|-3a + 0 - 1|}{\sqrt{(-3)^2 + 2^2}} = \frac{| -3a - 1|}{\sqrt{13}} = \frac{3a + 1}{\sqrt{13}}$$ (again positive since $a > 0$). - The chord length $l$ is related to radius and distance by: $$l = 2\sqrt{r^2 - d^2} = \frac{4}{\sqrt{13}}$$ 4. **Set up the equation for chord length:** $$2\sqrt{r^2 - d^2} = \frac{4}{\sqrt{13}} \implies \sqrt{r^2 - d^2} = \frac{2}{\sqrt{13}}$$ Square both sides: $$r^2 - d^2 = \frac{4}{13}$$ 5. **Substitute $r$ and $d$ expressions:** $$\left(\frac{a+1}{\sqrt{2}}\right)^2 - \left(\frac{3a+1}{\sqrt{13}}\right)^2 = \frac{4}{13}$$ Simplify: $$\frac{(a+1)^2}{2} - \frac{(3a+1)^2}{13} = \frac{4}{13}$$ Multiply both sides by $26$ (LCM of 2 and 13): $$13(a+1)^2 - 2(3a+1)^2 = 8$$ 6. **Expand and simplify:** $$13(a^2 + 2a + 1) - 2(9a^2 + 6a + 1) = 8$$ $$13a^2 + 26a + 13 - 18a^2 - 12a - 2 = 8$$ $$-5a^2 + 14a + 11 = 8$$ $$-5a^2 + 14a + 3 = 0$$ Multiply by $-1$: $$5a^2 - 14a - 3 = 0$$ 7. **Solve quadratic for $a$:** $$a = \frac{14 \pm \sqrt{14^2 - 4 \cdot 5 \cdot (-3)}}{2 \cdot 5} = \frac{14 \pm \sqrt{196 + 60}}{10} = \frac{14 \pm \sqrt{256}}{10} = \frac{14 \pm 16}{10}$$ - $a = \frac{14 + 16}{10} = 3$ (positive, valid) - $a = \frac{14 - 16}{10} = -0.2$ (negative, discard) So, $a = 3$. 8. **Find radius $r$:** $$r = \frac{3 + 1}{\sqrt{2}} = \frac{4}{\sqrt{2}} = 2\sqrt{2}$$ 9. **Hyperbola parameters:** - Center of circle $C$ is at $(3,0)$, which is one focus of hyperbola $H$. - The transverse axis length of $H$ equals the diameter of $C$, so: $$2\alpha = 2r = 4\sqrt{2} \implies \alpha = 2\sqrt{2}$$ 10. **Find $eta^2$ using focus formula:** For hyperbola $\frac{x^2}{\alpha^2} - \frac{y^2}{\beta^2} = 1$, foci are at $(\pm c, 0)$ where $$c^2 = \alpha^2 + \beta^2$$ Given one focus is at $(3,0)$, so $c = 3$. Calculate $\alpha^2$: $$\alpha^2 = (2\sqrt{2})^2 = 8$$ Then: $$c^2 = 9 = \alpha^2 + \beta^2 = 8 + \beta^2 \implies \beta^2 = 1$$ 11. **Calculate $2\alpha^2 + 3\beta^2$:** $$2 \times 8 + 3 \times 1 = 16 + 3 = 19$$ **Final answer:** $$\boxed{19}$$