JEE Main 2026 — 2 April, Evening Shift — Question 45

Let $\mathrm{B}(\mathrm{h}, \mathrm{k})$ Equation of circle with AB as diameter $(x-h)(x-3)+(y-k)(y-0)=0$ $x^{2}+y^{2}-(h+3) x-k y+3 h=0$ Centre $\left(\frac{\mathrm{h}+3}{2}, \frac{\mathrm{k}}{2}\right)$ Circle touches internally $\therefore \mathrm{C}_{1} \mathrm{C}_{2}=|\mathrm{R}-\mathrm{r}|$ $\sqrt{\left(\frac{\mathrm{h}+3}{2}\right)^{2}+\left(\frac{\mathrm{k}}{2}\right)^{2}}=\left|6-\frac{1}{2}\left(\sqrt{(\mathrm{~h}-3)^{2}+\mathrm{k}^{2}}\right)\right|$ $\sqrt{(x+3)^{2}+y^{2}}+\sqrt{(x-3)^{2}+y^{2}}=12$ $2 a=12 \therefore a=6$ $(-3,0)$ and $(3,0)$ are foci $2 \mathrm{ae}=6$ $12 \mathrm{e}=6$ $\mathrm{e}=\frac{1}{2}$ $\therefore 72 \mathrm{e}^{2}=72\left(\frac{1}{4}\right)=18$