A) \[{{x}^{2}}+{{y}^{2}}=6\]
B) \[{{x}^{2}}+{{y}^{2}}=9\]
C) \[({{x}^{2}}+1)y'+xy=0\]
D) \[zy'+({{x}^{2}}+1)y=0\]
Correct Answer: B
Solution :
Given, \[3x+2y=26\] \[\Rightarrow \] \[y=-\frac{3}{2}x+13\] And \[6x+y=31\] \[\Rightarrow \] \[x=-\frac{1}{6}y+\frac{31}{6}\] \[\therefore \] \[r=-\sqrt{\left( \frac{-3}{2} \right)\left( \frac{-1}{6} \right)}\] \[\Rightarrow \] \[r=-\frac{1}{2}\]
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