question_answer

The locus of the centres of the circles which touch externally the circles \[{{x}^{2}}+{{y}^{2}}={{a}^{2}}\]and \[{{x}^{2}}+{{y}^{2}}=4ax\], will be

A)            \[12{{x}^{2}}-4{{y}^{2}}-24ax+9{{a}^{2}}=0\]

B)            \[12{{x}^{2}}+4{{y}^{2}}-24ax+9{{a}^{2}}=0\]

C)            \[12{{x}^{2}}-4{{y}^{2}}+24ax+9{{a}^{2}}=0\]

D)            \[12{{x}^{2}}+4{{y}^{2}}+24ax+9{{a}^{2}}=0\]

Correct Answer: A

Solution :

           Let\[C\equiv (h,\ k)\], radius \[=r\]                    Co-ordinates of \[A\equiv \left[ \frac{ah}{a+r},\ \frac{ak}{a+r} \right]\]                    Co-ordinates of \[B\equiv \left[ \frac{2ar+2ah}{2a+r},\ \frac{2ak}{2a+r} \right]\]                    Putting co-ordinates of A and B in \[{{S}_{1}},\ {{S}_{2}}\] respectively and eliminating r, we get the locus                    \[12{{x}^{2}}-4{{y}^{2}}-24ax+9{{a}^{2}}=0\].                    Aliter: Since it touches \[{{x}^{2}}+{{y}^{2}}={{a}^{2}}\]and\[{{x}^{2}}+{{y}^{2}}-4ax=0\], therefore                    \[r+a=\sqrt{{{h}^{2}}+{{k}^{2}}}\]                                                         ?.(i)                          \[r+2a=\sqrt{{{(h-2a)}^{2}}+{{k}^{2}}}\]                                             ?.(ii)                    From (i), putting the value of r in (ii), we get                    \[-a+\sqrt{{{h}^{2}}+{{k}^{2}}}+2a=\sqrt{{{(h-2a)}^{2}}+{{k}^{2}}}\]                    On simplification, we get the required locus.