The correct option is
B 8For the given equation,
x2+6x+y2=4, complete the square for
x by adding 9 to both sides of the equation.
Hence,
x2+6x++9+y2=13
i.e., (x+3)2+y2=13.
Since x and y are integers, (x+3)2 and y2 are squares of integers less than 13.
Therefore we get 2 cases:
CASE 1:
(x+3)2=4 and y2=9
Then x+3=±2 and y=±3
→x=−3±2 and y=±3
Hence ,we get 4 sets of values of (x,y) as {(−1,3);(−5,3);(−1,−3);(−5,−3)}
CASE 2:
(x+3)2=9 and y2=4
Then x+3=±3 and y=±2
→x=−3±3 and y=±2
Hence, we get 4 sets of values of (x,y) as {(0,2);(−6,2);(0,−2);(−6,−2)}
The number of ordered pairs of integers (x,y) satisfying the given equation are 8.