Circle And System Of Circles Question 216
Question: The equation of a circle touching the axes of coordinates and the line $ x\cos \alpha +y\sin \alpha =2 $ can be
Options:
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
Show Answer
Answer:
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 }} $
therefore $ g=\frac{2}{\sin \alpha +\cos \alpha \pm 1} $ . Similarly other options hold.
BETA
