Let the circle $C$ 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 $-3 x+2 y=1$. 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 ______

Solution

$r=\left|\frac{a+1}{\sqrt{2}}\right| \Rightarrow(a+1)^2=2 r^2$ Also $\left(\frac{3 a-1}{\sqrt{13}}\right)^2+\left(\frac{2}{\sqrt{13}}\right)^2=r^2$ $\begin{gathered} \Rightarrow\left(\frac{3 a-1}{\sqrt{13}}\right)^2+\frac{4}{13}=\frac{(a+1)^2}{2} \\ 5 a^2-14 a-3=0 \end{gathered}$ $\therefore \quad a=-\frac{1}{5}, 3$ $\begin{aligned} & \because \quad a \neq-\frac{1}{5} \Rightarrow \\ & \Rightarrow \quad r=2 \sqrt{2} \end{aligned}$ $\because$ One focus of $\frac{x^2}{\alpha^2}-\frac{y^2}{\beta^2}=1$ is $(3,0)$ $\begin{aligned} & \Rightarrow \alpha e=3 \text { and } 2 \alpha=4 \sqrt{2} \\ & \Rightarrow \alpha=2 \sqrt{2} \Rightarrow \alpha^2=8 \\ & \alpha^2\left[1+\frac{\beta^2}{\alpha^2}\right]=9 \\ & \alpha^2+\beta^2=9 \\ & \Rightarrow \beta^2=1 \\ & \therefore 2 \alpha^2+3 \beta^2=19\end{aligned}$

Hence, the correct answer is 19.