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 $\dfrac{4}{\sqrt{13}}$ along the line $-3x + 2y = 1$. Let $H$ be the hyperbola $\dfrac{x^2}{\alpha^2} - \dfrac{y^2}{\beta^2} = 1$, 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 _____.
Step-by-Step Solution
Key Concept: Use the tangent and chord-bisection conditions to find the centre $(a,0)$ of $C$; compute $r$ from the tangent distance; the focus of $H$ is $(a,0)$ and $2\alpha = 2r$ gives $\alpha$; then $\beta^2 = \alpha^2(e^2-1)$ where $e = a/(\alpha)$.
Centre $C(a,0)$, $a>0$. Tangent to $x-y+1=0$: $r=\tfrac{|a+1|}{\sqrt{2}}=\tfrac{a+1}{\sqrt{2}}$. Chord length $\tfrac{4}{\sqrt{13}}$ along $-3x+2y=1$: distance from $(a,0)$ to this line $= \tfrac{|3a+1|}{\sqrt{13}}$, half-chord $=\tfrac{2}{\sqrt{13}}$, so $r^2-\tfrac{(3a+1)^2}{13}=\tfrac{4}{13}$. Substituting $r^2=\tfrac{(a+1)^2}{2}$: $\tfrac{(a+1)^2}{2}-\tfrac{(3a+1)^2}{13}=\tfrac{4}{13}$, giving $5a^2-14a-3=0$, so $a=3$ (taking $a>0$). $r=\tfrac{4}{\sqrt{2}}=2\sqrt{2}$. Transverse axis $=2r=4\sqrt{2}$, so $\alpha=2\sqrt{2}$, $\alpha^2=8$. Focus at $(3,0)$: $\alpha e=3 \Rightarrow e=\tfrac{3}{2\sqrt{2}}$. $\beta^2=\alpha^2(e^2-1)=8\left(\tfrac{9}{8}-1\right)=1$. $2\alpha^2+3\beta^2=16+3=19$.
Correct Answer: 19