1. Interpret the two regions

We are given $S=\{(x,y)\in \mathbb N\times \mathbb N: 9(x-3)^2+16(y-4)^2\le 144\}$ and $T=\{(x,y)\in \mathbb R\times \mathbb R:(x-7)^2+(y-4)^2\le 36\}.$

We need the number of points in $S\cap T$.

Since $S\subset \mathbb N\times \mathbb N$, the intersection $S\cap T$ consists of natural-number lattice points satisfying both inequalities.

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2. Rewrite the inequalities in standard form

For $S$: $9(x-3)^2+16(y-4)^2\le 144$ Divide by $144$: $\frac{(x-3)^2}{16}+\frac{(y-4)^2}{9}\le 1.$

So $S$ is the set of natural-number points inside/on the ellipse centered at $(3,4)$ with semi-axes $4$ and $3$.

For $T$: $(x-7)^2+(y-4)^2\le 36,$ which is the disk centered at $(7,4)$ with radius $6$.

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3. Find possible natural values of $x$ from the ellipse

From $\frac{(x-3)^2}{16}+\frac{(y-4)^2}{9}\le 1,$ we must have $\frac{(x-3)^2}{16}\le 1 \implies (x-3)^2\le 16.$ Thus $-4\le x-3\le 4 \implies -1\le x\le 7.$ Since $x\in \mathbb N$, possible values are $x=1,2,3,4,5,6,7.$

Now also impose the circle condition. Since the circle is centered at $x=7$ with radius $6$, $(x-7)^2\le 36 \implies 1\le x\le 13,$ so all $x=1$ to $7$ are allowed in principle.

Hence we check each integer $x=1,2,3,4,5,6,7$.

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4. For each $x$, find allowed integer $y$ values

We need both $9(x-3)^2+16(y-4)^2\le 144 \quad \text{and} \quad (x-7)^2+(y-4)^2\le 36.$

Because $y\in \mathbb N$, we count natural-number $y$ satisfying both.

Let $k=(y-4)^2$.

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Case 1: $x=1$

Ellipse: $9(1-3)^2+16(y-4)^2\le 144$ $9\cdot 4+16(y-4)^2\le 144$ $36+16(y-4)^2\le 144$ $16(y-4)^2\le 108$ $(y-4)^2\le \frac{27}{4}=6.75.$

So $|y-4|\le 2$, hence $y=2,3,4,5,6.$

Circle: $(1-7)^2+(y-4)^2\le 36$ $36+(y-4)^2\le 36$ $(y-4)^2\le 0,$ so $y=4.$

Thus for $x=1$, count $=1$.

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Case 2: $x=2$

Ellipse: $9(2-3)^2+16(y-4)^2\le 144$ $9+16(y-4)^2\le 144$ $16(y-4)^2\le 135$ $(y-4)^2\le \frac{135}{16}=8.4375.$ So $|y-4|\le 2$, hence $y=2,3,4,5,6.$

Circle: $(2-7)^2+(y-4)^2\le 36$ $25+(y-4)^2\le 36$ $(y-4)^2\le 11.$ So $|y-4|\le 3$, giving $y=1,2,3,4,5,6,7.$

Common values: $y=2,3,4,5,6.$ Count $=5$.

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Case 3: $x=3$

Ellipse: $16(y-4)^2\le 144 \implies (y-4)^2\le 9,$ so $y=1,2,3,4,5,6,7.$

Circle: $(3-7)^2+(y-4)^2\le 36$ $16+(y-4)^2\le 36$ $(y-4)^2\le 20.$ So $|y-4|\le 4$, giving $y=1,2,3,4,5,6,7,8.$

Common values: $y=1,2,3,4,5,6,7.$ Count $=7$.

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Case 4: $x=4$

Ellipse: $9(4-3)^2+16(y-4)^2\le 144$ $9+16(y-4)^2\le 144$ $(y-4)^2\le \frac{135}{16},$ so $y=2,3,4,5,6.$

Circle: $(4-7)^2+(y-4)^2\le 36$ $9+(y-4)^2\le 36$ $(y-4)^2\le 27,$ so $|y-4|\le 5$, hence $y=1,2,3,4,5,6,7,8,9.$

Common values: $y=2,3,4,5,6.$ Count $=5$.

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Case 5: $x=5$

Ellipse: $9(5-3)^2+16(y-4)^2\le 144$ $36+16(y-4)^2\le 144$ $16(y-4)^2\le 108$ $(y-4)^2\le 6.75,$ so $y=2,3,4,5,6.$

Circle: $(5-7)^2+(y-4)^2\le 36$ $4+(y-4)^2\le 36$ $(y-4)^2\le 32,$ so $|y-4|\le 5$, hence $y=1,2,3,4,5,6,7,8,9.$

Common values: $y=2,3,4,5,6.$ Count $=5$.

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Case 6: $x=6$

Ellipse: $9(6-3)^2+16(y-4)^2\le 144$ $81+16(y-4)^2\le 144$ $16(y-4)^2\le 63$ $(y-4)^2\le \frac{63}{16}=3.9375.$ So $|y-4|\le 1$, hence $y=3,4,5.$

Circle: $(6-7)^2+(y-4)^2\le 36$ $1+(y-4)^2\le 36$ $(y-4)^2\le 35,$ so $|y-4|\le 5$, hence $y=1,2,3,4,5,6,7,8,9.$

Common values: $y=3,4,5.$ Count $=3$.

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Case 7: $x=7$

Ellipse: $9(7-3)^2+16(y-4)^2\le 144$ $144+16(y-4)^2\le 144$ $(y-4)^2\le 0,$ so $y=4.$

Circle: $(7-7)^2+(y-4)^2\le 36$ $(y-4)^2\le 36,$ so $y$ can be many values, but intersection gives only $y=4.$

Count $=1$.

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5. Add all counts

Total number of points is $1+5+7+5+5+3+1=27.$

Therefore, $n(S\cap T)=27.$

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6. Comparison with stored answer

Stored correct answer: $27$

Our derived answer is also $27$, so it agrees.