A) \[{{x}^{2}}+{{y}^{2}}=1\]
B) \[{{x}^{2}}+{{y}^{2}}+2ax=0\]
C) \[{{x}^{2}}+{{y}^{2}}-2ax=0\]
D) \[{{x}^{2}}+{{y}^{2}}=2{{a}^{2}}\]
Correct Answer: C
Solution :
Equation of the circle which passes through origin is \[{{x}^{2}}+{{y}^{2}}+2gx+2fy=0\]. Radical axis with both circles is \[2gx+2fy+{{a}^{2}}=0\] ?.(i) \[2(g-a)x+2fy+2{{a}^{2}}=0\] ?.(ii) Also radical axis of the two circles is \[x=\frac{a}{2}\Rightarrow f=0\] From (i) and (ii), we get \[\frac{2g}{2(g-a)}=\frac{1}{2}\Rightarrow g=-a\] Hence circle is \[{{x}^{2}}+{{y}^{2}}-2ax=0\].
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