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 centre of the circle is, \[C=(h,k)\] and 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, \[12{{x}^{2}}-4{{y}^{2}}-24ax+9{{a}^{2}}=0\]