Question

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

• A. $2$
• B. $4$
• C. $6$
• D. $8$

Answer 1

The correct option is D

$8$

Explanation for the correct option:

Find the required number of ordered pairs.

An equation $x^2+6x+y^2=0$ is given.

Rewrite the given equation as follows:

$$\begin{gathered} x^2+6x+y^2=4 \\ \Rightarrow x^2+6x+9+y^2=4+9 \\ \Rightarrow {\left(x+3\right)}^2+y^2=13 \end{gathered}$$

Assume that, $m=x+3$ and $n=y$.

Therefore, $m^2+n^2=13$

Since, $x,y$ are integers. so, $m,n$ are also integers.

Which is possible only when $m=\pm 2,n=\pm 3$ or $m=\pm 3,n=\pm 2$.

When $m=\pm 2,n=\pm 3$.

Since, $m,n$ have two possible values each. So, $x,y$ also have two possible values each.

Therefore, the number of ordered pairs is $4$.

When $m=\pm 3,n=\pm 2$.

Since, $m,n$ have two possible values each. So, $x,y$ also have two possible values each.

Therefore, the number of ordered pairs is $4$.

Thus, the total number of ordered pairs is $4+4=8$.

Hence, option $D$ is the correct option.