question_answer

    The equation of a circle touching the axes of coordinates and the line\[x\text{ }cos\,\alpha +y\text{ }sin\,\alpha =2\] can be

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.