A) \[{{x}^{2}}+{{y}^{2}}+x+2y=0\]
B) \[{{x}^{2}}+{{y}^{2}}-x+20=0\]
C) \[{{x}^{2}}+{{y}^{2}}-x-2y=0\]
D) \[2({{x}^{2}}+{{y}^{2}})-x-2y=0\]
Correct Answer: C
Solution :
Family of circles is \[{{x}^{2}}+{{y}^{2}}-2x-4y+1+\lambda ({{x}^{2}}+{{y}^{2}}-1)=0\]. \[(1+\lambda ){{x}^{2}}+(1+\lambda ){{y}^{2}}-2x-4y+(1-\lambda )=0\] \[{{x}^{2}}+{{y}^{2}}-\frac{2}{1+\lambda }x-\frac{4}{1+\lambda }y+\frac{1-\lambda }{1+\lambda }=0\] ..... (i) Centre is \[\left[ \frac{1}{1+\lambda },\ \frac{2}{1+\lambda } \right]\] and radius \[=\sqrt{{{\left( \frac{1}{1+\lambda } \right)}^{2}}+{{\left( \frac{2}{1+\lambda } \right)}^{2}}-\frac{1-\lambda }{1+\lambda }}=\frac{\sqrt{4+{{\lambda }^{2}}}}{1+\lambda }\]. Since it touches the line\[x+2y=0\], hence Radius = Perpendicular from centre to the line i.e., \[\left| \frac{\frac{1}{1+\lambda }+2\frac{2}{1+\lambda }}{\sqrt{{{1}^{2}}+{{2}^{2}}}} \right|=\frac{\sqrt{4+{{\lambda }^{2}}}}{1+\lambda }\Rightarrow \sqrt{5}=\sqrt{4+{{\lambda }^{2}}}\Rightarrow \lambda \pm 1\] \[\Rightarrow \sqrt{5}=\sqrt{4+{{\lambda }^{2}}}\Rightarrow \lambda =\pm 1\] \[\lambda =-1\] cannot be possible in case of circle. So \[\lambda =1\]. Thus, from (i) \[{{x}^{2}}+{{y}^{2}}-x-2y=0\] is the required equation of the circle.
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