A) x-axis
B) y-axis
C) both axis
D) neither .y-axis nor y-axis
Correct Answer: B
Solution :
We have, \[{{x}^{2}}+{{y}^{2}}-8x+4y+4=0\] On comparing with standard equation of circle \[{{x}^{2}}+{{y}^{2}}+2gx+2fy+c=0\] , we get \[g=-4,f=2\] and \[c=4\] \[-g=4,\,-f=-2\] \[\therefore \] co-ordinate of the centre \[=(-g,\,-f)\] \[=\left( 4,-2 \right)\] \[\therefore \] Radius of the circle \[=\sqrt{{{g}^{2}}+{{f}^{2}}-{{c}^{2}}}\] \[=\sqrt{{{(-4)}^{2}}+{{(2)}^{2}}-4}\] \[=\sqrt{16+4-4}=4\] Here, radius of circle is equal to \[x\]-coordinate of the centre, \[\therefore \] circle touches y-axis
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