This equation can be written as $(x + 3)^{2} + y^{2} = 13.$ The only two perfect squares that add to $13$ are $4$ and $9$ , so we can have either

(i) $x + 3 = \pm 2$ and $y = \pm 3$ , or

(ii) $x + 3 = \pm 3$ and $y = \pm 2$ .

Each of these cases yield $2*2 = 4$ distinct ordered pairs, giving a final solution of $\boxed{8}$ ordered pairs.

For completeness, the $8$ ordered pairs $(x,y)$ are

$(-5,-3), (-5,3), (-1, -3), (-1,3), (-6,-2), (-6,2), (0,-2), (0,2).$

Rewrite the equation as:

$x^2 + 6x + (y^2 - 4) = 0$

Solving for x gives us:

$x = \frac{-6 \pm \sqrt{36 - 4(y^2 - 4)}}{2} = -3 \pm \sqrt{13 - y^2}$

It can be seen that $|y| < 4$ (since we don't want any imaginary numbers). Also, $y = 0$ and $y = \pm1$ give non-integer solutions for x.

Hence the integer solutions are:

$(-6,\pm2), (0,\pm2), (-5,\pm3), (-1,\pm3)$

Imgur

For a circle with center at the origin, reflection of a point with integer coordinates about X and Y axis will give us three more points with integer coordinates. More over a point (h,k) reflected about y = x will give another point with integer coordinates (k,h).

Hence we need to inspect only $\frac{1}{8}$ of a circle for such points. On such an arc from 0 to 45 degrees, all points will have x > y AND x would lie between $\frac{R}{\sqrt{2}}$ and $int(R)$

Here these two limits leave only one value of x viz. 3. So we just check that:

For the given circle with center (-3,0) check x = -3 + 3 = 0, which gives y = 2

Since we have just one point in the $\frac{1}{8}$ arc, we will have 1 x 8 = 8 such points.

@Niranjan Khanderia

include <iostream>

using namespace std;

int main() { int x,y,num;

for( x=-8 ; x<2 ; x++ )
    for( y=-4 ; y<5 ; y++ )
    {
        num = (x+3)*(x+3) + y*y;
        if( num==13 )
            cout<<x<<" "<<y<<"\n";
    }
return 0;

}

Output: no. of ordered pairs (x,y) that satisfies the equation