A) \[{{x}^{2}}+{{y}^{2}}+17x+19y-50=0\]
B) \[{{x}^{2}}+{{y}^{2}}-17x-19y-50=0\]
C) \[{{x}^{2}}+{{y}^{2}}+17x-19y-50=0\]
D) \[{{x}^{2}}+{{y}^{2}}-17x+19y+50=0\]
E) \[{{x}^{2}}+{{y}^{2}}-17x\,+19y+50=0\]
Correct Answer: D
Solution :
Lines, \[x+y=6,2x+y=4\]and\[x+2y=5\] Intersect at points\[(-2,8),(7-1)\]and (1, 2) Now, all these points lie on\[{{x}^{2}}+{{y}^{2}}-17x-19y+50=0\].
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