JEE Main 2026 — 23 January, Evening Shift — Question 8

By family of curve equation of circle will be $\Rightarrow \mathrm{S}_{1}+\lambda \mathrm{S}_{2}=0$

$\Rightarrow\left(\mathrm{x}^{2}+2 \mathrm{y}^{2}-6 \mathrm{x}-12 \mathrm{y}+23\right)$

$+\lambda\left(4 x^{2}+2 y^{2}-20 x-12 y+35\right)=0$ ⇒ for circle coeff of $x^{2}=$ coeff. of $y^{2}$

$\Rightarrow \lambda=\frac{1}{2}$

So equation of circle is $\Rightarrow \mathrm{x}^{2}+\mathrm{y}^{2}-\frac{16}{3} \mathrm{x}-6 \mathrm{y}+\frac{27}{2}=0$

Centre $\left(\frac{8}{3}, 3\right):$ Radius $\mathrm{r}=\sqrt{\frac{47}{18}}=\mathrm{r}$

$\therefore \mathrm{ab}+18 \mathrm{r}^{2}=8+47=55$