A) - 2
B) 2
C) 1
D) 0
Correct Answer: B
Solution :
The equation of circle passing through \[(0,\,0),\,\,(2,\,0)\] and (0, ? 2) is \[{{x}^{2}}+{{y}^{2}}-2x+2y=0\]. If it passes through \[(k,\,-2)\], then \[{{k}^{2}}+4-2k-4=0\]\[\Rightarrow \,k=0,\,2\] \[\because \]\[(0,\,-2)\] is already a point on circle \[\therefore \] k = 2.You need to login to perform this action.
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