JEE Main 2025 — 4 April, Morning Shift — Question 38

$A=\left\{(x, y) \in R \times R: x^{2}+y^{2}=25\right\}, B=\{(x, y) \in \mathbb{R} \times$

$\left.\mathbb{R}: x^{2}+9 y^{2}=144\right\}$

$x^{2}+9 y^{2}-\left(x^{2}+y^{2}\right)=144-25$

Plug in $y^{2}=\frac{119}{8}$ into either equation to find $x$. $x^{2}=25-\frac{119}{8}$ $x^{2}=\frac{200-119}{8}$

$x^{2}=\frac{81}{8}$

$x= \pm \sqrt{\frac{81}{8}}, y= \pm \sqrt{\frac{119}{8}}$

Now, $C=\left\{(x, y) \in \mathbb{Z} \times \mathbb{Z}: x^{2}+y^{2} \leq 4\right\}$

Valid points are $(-2,0),(-1,-1),(-1,0),(-1,1)$,

$(0,-2),(0,-1),(0,0),(0,1),(0,2),(1,-1),(1,0)$,

$(1,1)$

$\therefore$ Total valid points in $C=13$

$\Rightarrow$ There are 4 distinct real points in set $D$

$\therefore$ The number of one-one functions from $D$ to $C$

$\Rightarrow \quad 13 P_{4} \Rightarrow \frac{13!}{(13-4)!}=\frac{13!}{9!}=17160$