A) \[{{x}^{2}}+{{y}^{2}}-2x+6y-22=0\]
B) \[{{x}^{2}}+{{y}^{2}}+2x+6y-11=0\]
C) \[{{x}^{2}}+{{y}^{2}}-2x+6y+22=0\]
D) \[{{x}^{2}}+{{y}^{2}}-2x-6y-22=0\]
Correct Answer: D
Solution :
Since, \[\angle APB=\frac{\pi }{2}\] Therefore, A and B are the extreme points of the diameter of the circle. We know that, equation of circle with \[({{x}_{1}},{{y}_{1}})\]and \[({{x}_{2}},{{y}_{2}})\] be the end points of diameter is \[(x-{{x}_{1}})\,(x-{{x}_{2}})+(y-{{y}_{1}})\,(y-{{y}_{2}})=0\] Therefore, equation of circle with \[(5,-1)\] and \[(-3,7)\] be the end points of diameter is \[(x-5)\,(x+3)+(y+1)\,(y-7)=0\] \[\Rightarrow \] \[{{x}^{2}}-2x-15+{{y}^{2}}-6y-7=0\] \[\Rightarrow \] \[{{x}^{2}}+{{y}^{2}}-2x-6y-22=0\] Which is the required equation of circle.You need to login to perform this action.
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