1. Find the set $D=A\cap B$

We need points $(x,y)$ satisfying both $x^2+y^2=25 \quad \text{and} \quad x^2+9y^2=144.$

Subtract the first equation from the second: $x^2+9y^2-(x^2+y^2)=144-25$ $8y^2=119$ $y^2=\frac{119}{8}.$

Then $x^2=25-y^2=25-\frac{119}{8}=\frac{81}{8}.$

So the common points are $(x,y)=\left(\pm \frac{9}{2\sqrt2}, \pm \sqrt{\frac{119}{8}}\right),$ with independent signs. Hence there are $|D|=4$ points.

2. Find the set $C$

$C=\{(x,y)\in \mathbb Z\times \mathbb Z: x^2+y^2\le 4\}.$

We count all integer lattice points inside or on the circle of radius $2$.

Possible integer values:

• If $x=0$, then $y^2\le 4 \Rightarrow y=0,\pm1,\pm2$ : $5$ points.
• If $x=\pm1$, then $1+y^2\le 4 \Rightarrow y^2\le 3 \Rightarrow y=0,\pm1$ : $3$ points for each of $x=1,-1$, total $6$.
• If $x=\pm2$, then $4+y^2\le 4 \Rightarrow y=0$ : $1$ point for each, total $2$.

Thus $|C|=5+6+2=13.$

3. Count one-one functions from $D$ to $C$

Since $|D|=4$ and $|C|=13$, the number of injective functions from $D$ to $C$ is $^{13}P_4=13\cdot 12\cdot 11\cdot 10.$

Compute: $13\cdot 12=156,$ $11\cdot 10=110,$ $156\cdot 110=17160.$

So the total number of one-one functions is $\boxed{17160}.$

4. Compare with stored answer

Our derived answer is Option C: 17160, which matches the stored correct answer.