QPaperGen

We are given:
\[
A = \{(x, y) \in \mathbb{R}^2 : x^2 + y^2 = 25\}, \quad
B = \{(x, y) \in \mathbb{R}^2 : x^2 + 9y^2 = 144\}.
\]

Subtracting the two equations:
\[
x^2 + 9y^2 - (x^2 + y^2) = 144 - 25 \quad \Rightarrow \quad 8y^2 = 119
\]
\[
\Rightarrow \quad y^2 = \tfrac{119}{8}.
\]

Substitute into \(x^2 + y^2 = 25\):
\[
x^2 = 25 - \tfrac{119}{8} = \tfrac{200 - 119}{8} = \tfrac{81}{8}.
\]

Hence,
\[
x = \pm \sqrt{\tfrac{81}{8}}, \quad y = \pm \sqrt{\tfrac{119}{8}}.
\]

So, the set \(D = A \cap B\) has **4 distinct real points**.

---

Now, for set \(C\):
\[
C = \{(x, y) \in \mathbb{Z}^2 : x^2 + y^2 \leq 4\}.
\]

Valid integer points are:
\[
(-2,0), \; (-1,-1), \; (-1,0), \; (-1,1), \; (0,-2), \; (0,-1), \; (0,0), \; (0,1), \; (0,2), \; (1,-1), \; (1,0), \; (1,1), \; (2,0).
\]

Thus, \(|C| = 13\).

---

We need the number of one-one functions from \(D\) (with \(|D|=4\)) to \(C\) (with \(|C|=13\)):
\[
= {}^{13}P_{4} = \frac{13!}{(13-4)!} = \frac{13!}{9!}.
\]

\[
= 13 \times 12 \times 11 \times 10 = 17160.
\]

\[
\therefore \quad \text{The total number of one-one functions is } 17160.
\]