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) \[{{x}^{2}}+{{y}^{2}}-4x-4y-4=0\]
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 \[\left| \frac{3(-a)-4(-a)+8}{\sqrt{9+16}} \right|\ =a\Rightarrow a=2\] 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).
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