Question

The number of ordered pairs of integers $(x,y)$ satisfying the equation $x^2+6x+y^2=4$ is

• A. $2$
• B. $8$
• C. $6$
• D. None of these

Answer 1

The correct option is B $8$

For the given equation, $x^2+6x+y^2=4$, complete the square for $x$ by adding 9 to both sides of the equation.
Hence,

$x^2+6x++9+y^2=13$
i.e., ${(x+3)}^2+y^2=13$.

Since $x$ and $y$ are integers, ${(x+3)}^2$ and $y^2$ are squares of integers less than $13$.

Therefore we get 2 cases:
CASE 1:

${(x+3)}^2=4$ and $y^2=9$
Then $x+3=\pm 2$ and $y=\pm 3$
$\to x=-3\pm 2$ and $y=\pm 3$
Hence ,we get 4 sets of values of $(x,y)$ as $\left\{\left(-1,3\right);\left(-5,3\right);\left(-1,-3\right);\left(-5,-3\right)\right\}$
CASE 2:

${(x+3)}^2=9$ and $y^2=4$
Then $x+3=\pm 3$ and $y=\pm 2$
$\to x=-3\pm 3$ and $y=\pm 2$
Hence, we get 4 sets of values of $(x,y)$ as $\left\{\left(0,2\right);\left(-6,2\right);\left(0,-2\right);\left(-6,-2\right)\right\}$

The number of ordered pairs of integers $(x,y)$ satisfying the given equation are $8$.