A) \[{{x}^{2}}+{{y}^{2}}+ax=0\]
B) \[{{x}^{2}}+{{y}^{2}}-ax=0\]
C) \[{{x}^{2}}+{{y}^{2}}+ay=0\]
D) \[{{x}^{2}}+{{y}^{2}}-ay=0\]
Correct Answer: B
Solution :
The given circle \[{{x}^{2}}+{{y}^{2}}-2ax=0\] ...(i) Let centre of the drawn circle be\[(\alpha ,\beta )\]let variable chord \[y=mx\] ...(ii) On solving Eqs. (i) and (ii), we get \[x=0,\frac{2a}{1+{{m}^{2}}}\] and \[y=0,\frac{2am}{1+{{m}^{2}}}\] \[\therefore \]Intersection points are\[(0,0)\left( \frac{2a}{1+{{m}^{2}}},\frac{2am}{1+{{m}^{2}}} \right)\] Let coordinate of the centre of circle be\[(\alpha ,\beta )\] \[\therefore \] \[\alpha =\frac{0+\frac{2a}{1+{{m}^{2}}}}{2}=\frac{a}{1+{{m}^{2}}},\] and \[\beta =\frac{0+\frac{2am}{1+{{m}^{2}}}}{2}=\frac{am}{1+{{m}^{2}}},\] Thus, \[\beta =\alpha m\] \[\Rightarrow \] \[m=\frac{\beta }{\alpha }\] Then, \[\alpha =\frac{a}{1+\frac{{{\beta }^{2}}}{{{\alpha }^{2}}}}\] \[\Rightarrow \] \[\alpha =\frac{a{{\alpha }^{2}}}{{{\alpha }^{2}}+{{\beta }^{2}}}\] \[\Rightarrow \] \[\alpha ({{\alpha }^{2}}+{{\beta }^{2}})=a{{\alpha }^{2}}\] \[\Rightarrow \] \[{{\alpha }^{2}}+{{\beta }^{2}}-a\alpha =0\] Hence, locus of the centre of circle is \[{{x}^{2}}+{{y}^{2}}-ax=0\]
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