A) \[{{x}^{2}}+{{y}^{2}}-6x+7=0\]
B) \[{{x}^{2}}+{{y}^{2}}-3y+4=0\]
C) \[{{x}^{2}}+{{y}^{2}}-2x-2y+1=0\]
D) \[{{x}^{2}}+{{y}^{2}}+2x-4y+4=0\]
Correct Answer: A
Solution :
Required circle will be\[{{S}_{1}}+\lambda {{S}_{2}}=0\], \[\lambda \ne -1\] i.e., \[{{x}^{2}}+{{y}^{2}}-2x-4y+1+\lambda ({{x}^{2}}+{{y}^{2}}-4x-2y+4)=0\] Þ \[{{x}^{2}}+{{y}^{2}}-2\frac{(1+2\lambda )}{1+\lambda }x-2\frac{(2+\lambda )}{1+\lambda }y+\frac{1+4\lambda }{1+\lambda }=0\] Its centre \[\left( \frac{1+2\lambda }{1+\lambda },\ \frac{2+\lambda }{1+\lambda } \right)\] lies on \[x+2y-3=0\] \ \[\frac{1+2\lambda }{1+\lambda }+2\left( \frac{2+\lambda }{1+\lambda } \right)-3=0\] Þ \[\lambda =-2\]. \ The circle is \[{{x}^{2}}+{{y}^{2}}-6x+7=0\]
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