JEE Main 2025 Jan 23 Shift 1, Mathematics Q23: Circle: Chord, Tangent And Normal

Let the centre be $({\alpha }_0,0)$ with ${\alpha }_0>0$, and radius $r$. Tangency to $x-y+1=0$ gives $\frac{|{\alpha }_0+1|}{\sqrt{2}}=r$, so $({\alpha }_0+1{)}^2=2r^2$ ... (1).

For the chord on $-3x+2y=1$: distance from centre to this line is $\frac{|-3{\alpha }_0-1|}{\sqrt{13}}$, and the chord-length relation ${\text{(half chord)}}^2+d^2=r^2$ gives $\frac{(3{\alpha }_0+1{)}^2}{13}+\frac{4}{13}=r^2$, i.e. $(3{\alpha }_0+1{)}^2+4=13r^2$ ... (2).

From (1), $r^2=\frac{({\alpha }_0+1{)}^2}{2}$. Substituting into (2): $2[(3{\alpha }_0+1{)}^2+4]=13({\alpha }_0+1{)}^2$, expanding gives $5{\alpha }_0^2-14{\alpha }_0-3=0$, so ${\alpha }_0=3$ or ${\alpha }_0=-1/5$. Positive root: ${\alpha }_0=3$. Then $r=\frac{4}{\sqrt{2}}=2\sqrt{2}$.

Hyperbola: transverse-axis length $=2r$ gives semi-axis $a=r=2\sqrt{2}$ (call this $\alpha$ for the question), so ${\alpha }^2=8$. Focus at $({\alpha }_0,0)=(3,0)$ means $ae=3$, so $e=\frac{3}{2\sqrt{2}}$, and ${\beta }^2=a^2(e^2-1)=8\left(\frac{9}{8}-1\right)=1$.

$2{\alpha }^2+3{\beta }^2=16+3=19$.