Circle And System Of Circles Question 158
Question: The equation of the circle which passes through the origin and cuts off intercepts of 2 units length from negative coordinate axes, is
Options:
A) $ x^{2}+y^{2}-2x+2y=0 $
B) $ x^{2}+y^{2}+2x-2y=0 $
C) $ x^{2}+y^{2}+2x+2y=0 $
D) $ x^{2}+y^{2}-2x-2y=0 $
Show Answer
Answer:
Correct Answer: C
Solution:
Since the circle passes through (0, 0),
Hence $ c=0 $ . Also $ 2\sqrt{g^{2}-c}=2\Rightarrow g=1 $ and $ 2\sqrt{f^{2}-c}=2\Rightarrow f=1 $ .
Hence radius is $ \sqrt{2} $ and centre is $ (-1,\ -1) $ .
Therefore, the required equation is $ x^{2}+y^{2}+2x+2y=0 $ .
Trick: Obviously the centre of circle lies in III quadrant, which is given by (c).
BETA
