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 $-3x+2y=1$. 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
Solution:
Let the center of circle $C$ be $(h, 0)$ with $h > 0$.
The radius $r$ is the perpendicular distance from $(h, 0)$ to $x-y+1=0$:
$$r = \left| \frac{h-0+1}{\sqrt{1^2+(-1)^2}} \right| = \frac{|h+1|}{\sqrt{2}} \quad \dots(1)$$
Distance $d$ from center $(h, 0)$ to the chord line $3x-2y+1=0$:
$$d = \left| \frac{3(h)-2(0)+1}{\sqrt{3^2+(-2)^2}} \right| = \frac{|3h+1|}{\sqrt{13}}$$
Length of chord $L = \frac{4}{\sqrt{13}}$, so semi-chord length is $\frac{2}{\sqrt{13}}$.
In the right triangle formed by radius, distance $d$, and semi-chord:
$$r^2 = d^2 + \left(\frac{2}{\sqrt{13}}\right)^2$$
Substitute values:
$$\frac{(h+1)^2}{2} = \frac{(3h+1)^2}{13} + \frac{4}{13}$$
$$13(h^2+2h+1) = 2(9h^2+6h+1) + 8$$
$$13h^2+26h+13 = 18h^2+12h+10$$
$$5h^2 – 14h – 3 = 0 \Rightarrow (5h+1)(h-3) = 0$$
Since $h > 0$, we get $h=3$.
Radius $r = \frac{|3+1|}{\sqrt{2}} = 2\sqrt{2}$. Diameter $= 4\sqrt{2}$.
$\Rightarrow$ For Hyperbola $\frac{x^2}{\alpha^2} – \frac{y^2}{\beta^2} = 1$:
1. Transverse axis ($2\alpha$) = Diameter of $C \Rightarrow 2\alpha = 4\sqrt{2} \Rightarrow \alpha = 2\sqrt{2} \Rightarrow \alpha^2 = 8$.
2. Focus ($ae$) = Center of $C \Rightarrow \alpha e = 3$.
3. Relation $b^2 = a^2(e^2-1) \Rightarrow \beta^2 = (\alpha e)^2 – \alpha^2$.
$$\beta^2 = 3^2 – 8 = 1$$
Value of $2\alpha^2 + 3\beta^2$:
$$2(8) + 3(1) = 16 + 3 = 19$$
Ans. 19