A) \[{{x}^{2}}+{{y}^{2}}-2gx-2gy+{{g}^{2}}=0\], where \[g=\frac{2}{(\cos \alpha +\sin \alpha +1)}\]
B) \[{{x}^{2}}+{{y}^{2}}-2gx-2gy+{{g}^{2}}=0\], where \[g=\frac{2}{(\cos \alpha +\sin \alpha -1)}\]
C) \[{{x}^{2}}+{{y}^{2}}-2gx+2gy+{{g}^{2}}=0\], where \[g=\frac{2}{(\cos \alpha -\sin \alpha +1)}\]
D) \[{{x}^{2}}+{{y}^{2}}-2gx+2gy+{{g}^{2}}=0\] where \[g=\frac{2}{(\cos \alpha +\sin \alpha +1)}\]
E) All of these
Correct Answer: E
Solution :
\[{{x}^{2}}+{{y}^{2}}-2gx-2gy+{{g}^{2}}=0\] \[g=\pm \frac{g\cos \alpha +g\sin \alpha -2}{\sqrt{{{\sin }^{2}}\alpha +{{\cos }^{2}}\alpha }}\] Þ \[g=\frac{2}{\sin \alpha +\cos \alpha \pm 1}\]. Similarly other options hold.You need to login to perform this action.
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