A) \[2{{x}^{2}}+2{{y}^{2}}-4x+1=0\]
B) \[{{x}^{2}}+{{y}^{2}}-4x+2=0\]
C) \[{{x}^{2}}+{{y}^{2}}-4x+1=0\]
D) \[2{{x}^{2}}+2{{y}^{2}}-4x+3=0\]
Correct Answer: A
Solution :
Let \[P(\alpha ,\,\beta )\] be on the circle \[{{S}_{1}}:{{x}^{2}}+{{y}^{2}}-2x-1=0\] \[\therefore \,\,{{\alpha }^{2}}+{{\beta }^{2}}-2\alpha -1=0\] ? (i) Equation of AB is \[\alpha x+\beta y-(x+\alpha )=0\]\[(T=0)\] \[\Rightarrow (\alpha -1)x+\beta y-\alpha =0\] New equation of circle passing through the point of intersection at \[{{S}_{2}}=0\] and AB is \[{{x}^{2}}+{{y}^{2}}-2x+\lambda \,\left( (\alpha -1)x+\beta y-\alpha \right)=0\] This circle is passing through (1, 0) \[\therefore \,\,1-2+\lambda (\alpha -1-\alpha )=0\] \[\Rightarrow \,\,\lambda =-1\] \[\therefore \] Equation of circumcircle of \[\Delta CAB\] is \[{{x}^{2}}+{{y}^{2}}-(\alpha -1)x-\beta y+\alpha =0\] \[\therefore \] Coordinates of centre \[\left( \frac{\alpha +1}{2},\,\frac{\beta }{2} \right)\]. \[\therefore \] Let \[\frac{\alpha +1}{2}=h,\,\frac{\beta }{2}=k\] \[\Rightarrow \,\alpha =2h-1,\,\beta =2k\] \[\therefore \] from equation (i) \[{{(2h-1)}^{2}}+4{{k}^{2}}-2(2h-1)-1=0\] \[\therefore \] Locus of circumcircle \[2{{x}^{2}}+2{{y}^{2}}-4x+1=0\]
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