Q.

The locus of the centre of the circles which touch both the circles x2+y2=a2 and x2+y2=4ax externally has the equation

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Detailed Solution

If the circle (x−h)2+(y−k)2=r2 touches both the circle x2+y2=a2 and x2+y2−4ax=0 externally. Then,h2+k2=r+a and (h−2a)2+k2=r+2a∴    (h−2a)2+k2−h2+k2=a⇒    (h−2a)2+k2=a+h2+k2⇒    (h−2a)2+k2=a2+h2+k2+2ah2+k2⇒    −4ah+3a2=2ah2+k2⇒    (3a−4h)2=4h2+k2⇒12(h−a)2−4k2=3a2Hence, the locus of (h,k) is 12(x−a)2−4y2=3a2
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