A) \[\sqrt{26}\]
B) \[\sqrt{13}\]
C) \[\sqrt{23}\]
D) 0
Correct Answer: D
Solution :
Given equation is \[{{x}^{2}}+{{y}^{2}}+4x+6y+13=0\] or \[\left( {{x}^{2}}+4x+4 \right)+\left( {{y}^{2}}+6y+9 \right)+13=4+9\] or \[{{\left( x+2 \right)}^{2}}+{{\left( y+3 \right)}^{2}}=0\] \[\therefore \] radius of circle = 0.You need to login to perform this action.
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