Circle And System Of Circles Question 120
Question: The equation of a circle which touches both axes and the line $ 3x-4y+8=0 $ and whose centre lies in the third quadrant is
[MP PET 1986]
Options:
A) $ x^{2}+y^{2}-4x+4y-4=0 $
B) $ x^{2}+y^{2}-4x+4y+4=0 $
C) $ x^{2}+y^{2}+4x+4y+4=0 $
D) $ x^{2}+y^{2}-4x-4y-4=0 $
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Answer:
Correct Answer: C
Solution:
The equation of circle in third quadrant touching the coordinate axes with centre $ (-a,\ -a) $ and radius -a- is $ x^{2}+y^{2}+2ax+2ay+a^{2}=0 $ and we know $ | \frac{3(-a)-4(-a)+8}{\sqrt{9+16}} |\ =a\Rightarrow a=2 $
Hence the required equation is $ x^{2}+y^{2}+4x+4y+4=0 $ .
Trick: Obviously the centre of the circle lies in III quadrant, which is given by (c).
BETA
