A) \[{{x}^{2}}+{{y}^{2}}-8x=0\]
B) \[{{x}^{2}}+{{y}^{2}}-5x+7y=0\]
C) \[{{x}^{2}}+{{y}^{2}}-5x+7y-1=0\]
D) \[{{x}^{2}}+{{y}^{2}}-8x+7y-2=0\]
Correct Answer: A
Solution :
Let \[S={{x}^{2}}+{{y}^{2}}-8x\] At point \[(5,-7)\] \[S={{5}^{2}}+{{(-7)}^{2}}-8(5)\] \[=34>0\] So, point lies outside the circle. Hence, option [a] is correct.
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