A) \[{{x}^{2}}+{{y}^{2}}+\text{ }8x+2y+8=0\]
B) \[{{x}^{2}}+{{y}^{2}}+8x+2y-8=0\]
C) \[{{x}^{2}}+{{y}^{2}}+8x+2y-14=0\]
D) None of the above
Correct Answer: B
Solution :
The equation of given circle is \[{{x}^{2}}+{{y}^{2}}+14x+10y-26=0\] ?(i) Its centre\[(-g,-f)=(-7-5)\] and radius \[=\sqrt{{{g}^{2}}+{{f}^{2}}-c}\] \[=\sqrt{49+25+26}=10\] Also, the required circle is of radius 5 or diameter 10. Therefore, it passes through the centre of the circle (i). Thus, the points \[(-7,-5)\] and \[(-1,3)\] are the end points of the diameter of required circle which is given by \[(x-{{x}_{1}})(x-{{x}_{2}})+(y-{{y}_{1}})(y-{{y}_{2}})=0\] where \[({{x}_{1}},{{y}_{1}})=(-7,-5)\] and \[({{x}_{2}},{{y}_{2}})=(-1,3)\] \[\therefore \]Required circle is given by \[(x+7)(x+1)+(y+5)(y-3)=0\] \[\Rightarrow \]\[{{x}^{2}}+8x+7+{{y}^{2}}+2y-15=0\] \[\Rightarrow \]\[{{x}^{2}}+{{y}^{2}}+8x+2y-8=0\]You need to login to perform this action.
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