Let A(-4, 0) & B(4,0). Then the number of points C=(x,y) on the circle ${x^2} + {y^2} = 16$ lying in first quadrant $(x,y \geqslant 0)$ such that the area of the triangle whose vertices are A,B,C is a integer is 

$\begin{array}{l} A(-4,0) ; B(4,0) \\ x^{2}+y^{2}=16 \end{array}$

centre $(0,0)$

$\text { radius }=4$

$\begin{aligned} \text { Area } & =\frac{1}{2} \times 8 \times 4 \sin \theta \\ & =16 \sin \theta \\ \sin \theta & =\frac{1}{16}, \frac{2}{16}, \ldots . \frac{15}{16} \rightarrow 15 \text { Points } \end{aligned}$

$\therefore$ There are 15 points for each vuadrant.