A) \[{{x}^{2}}+{{y}^{2}}+x+2y=0\]
B) \[{{x}^{2}}+{{y}^{2}}-x+2y=0\]
C) \[{{x}^{2}}+{{y}^{2}}+x-2y=0\]
D) \[{{x}^{2}}+{{y}^{2}}-x-2y=0\]
Correct Answer: D
Solution :
[d]: The equation of the circle passing through the intersection of the circle \[{{x}^{2}}+{{y}^{2}}=4\]and \[{{x}^{2}}+{{y}^{2}}-2x-4y+4=0\] is \[({{x}^{2}}+{{y}^{2}}-2x-4y+4)+\lambda ({{x}^{2}}+{{y}^{2}}-4)=0\] \[\Rightarrow \]\[(1+\lambda ){{x}^{2}}+(1+\lambda ){{y}^{2}}-2x-4y+4(1-\lambda )=0\] \[\Rightarrow \]\[{{x}^{2}}+{{y}^{2}}-\frac{2x}{1+\lambda }-\frac{4y}{1+\lambda }+\frac{4(1-\lambda )}{1+\lambda }=0\] ?(i) Since the line \[x+2y=0\]touches the circle (i) \[\therefore \]\[\frac{\frac{1}{1+\lambda }+\frac{1}{1+\lambda }}{\sqrt{{{1}^{2}}+{{2}^{2}}}}=\sqrt{{{\left( \frac{1}{1+\lambda } \right)}^{2}}+{{\left( \frac{2}{1+\lambda } \right)}^{2}}-\frac{4(1-\lambda )}{1+\lambda }}\] \[\Rightarrow \]\[\frac{\frac{5}{1+\lambda }}{\sqrt{5}}=\sqrt{\frac{5-(4)(1-{{\lambda }^{2}})}{{{(1+\lambda )}^{2}}}}\] \[\Rightarrow \]\[\frac{\sqrt{5}}{1+\lambda }=\sqrt{\frac{1+4{{\lambda }^{2}}}{{{(1+\lambda )}^{2}}}}\Rightarrow \frac{5}{{{(1+\lambda )}^{2}}}=\frac{1+4{{\lambda }^{2}}}{{{(1+\lambda )}^{2}}}\] \[1+4{{\lambda }^{2}}=5\Rightarrow {{\lambda }^{2}}=1\Rightarrow \lambda =1\] Now put the value of \[\lambda =1\]in equation (i) we get \[{{x}^{2}}+{{y}^{2}}-x-2y=0\]
You need to login to perform this action.
You will be redirected in
3 sec
