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) All of the above
Correct Answer: D
Solution :
\[{{x}^{2}}+{{y}^{2}}-2gx-2gy+{{g}^{2}}=0\] \[\therefore \] \[g=\pm \frac{g\cos \alpha +g\sin \alpha -2}{\sqrt{{{\sin }^{2}}\alpha +{{\cos }^{2}}\alpha }}\] \[=\frac{2}{\sin \alpha +\cos \alpha \pm 1}\] Similarly, other option hold.You need to login to perform this action.
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