A) \[7{{x}^{2}}+7{{y}^{2}}-10x+10y-11=0\]
B) \[7{{x}^{2}}+7{{y}^{2}}+10x-10y-12=0\]
C) \[7{{x}^{2}}+7{{y}^{2}}-10x-10y-12=0\]
D) \[7{{x}^{2}}+7{{y}^{2}}-10x-12=0\]
Correct Answer: C
Solution :
Family of circles through points of intersection of two circles is\[{{S}_{1}}+\lambda {{S}_{2}},\,\,(\lambda \ne -1)\]. \[{{x}^{2}}+{{y}^{2}}-6x+2y+4+\lambda ({{x}^{2}}+{{y}^{2}}+2x-4y-6)=0\] Centre is\[(3-\lambda ,\ -1+2\lambda )\]. It lies on\[y=x\]. Therefore, \[-1+2\lambda =3-\lambda \Rightarrow \lambda =\frac{4}{3}\] Hence equation of circle can be found by substituting \[\lambda \] in the family of circles above.
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