Two identical coherent sources are placed on a diameter of a circle of radius R at separation x(<<R) symmetrical about the center of the circle. The sources emit identical wavelength $\lambda$ each. The number of options on the circle of maximum intensity us $\left(x=5\lambda \right)$

Path difference at P is

$\Delta x=2\left(\frac{x}{2}\cos \theta \right)=xcos\theta$

 for intensity to be maximum

$\Delta x=n\lambda$     $\left(n=0,1,2,3,\mathrm{........}\right)$

Or $\cos \theta =\frac{n\lambda }{x}\ge 1$

$\therefore n\ge \frac{x}{\lambda }$

Substituting $x=5\lambda$ , we get

$n\ge 5$ or $n=1,2,3,4,5,\mathrm{........}$

Therefore, in all four quadrants there can be 20 maxima. These are more maxima at $\theta =0^0$ and$\theta ={180}^0$ . But n=5 corresponds to $\theta ={90}^0$ and $\theta ={270}^0$ which are coming only twice while we have multiplied it four times. Therefore total number of maxima is still 20. I.e. n=1 to 4 in four quadrants (total) 16 plus more at $\theta =0^0,{90}^0,{180}^0$ and ${270}^0$ .