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Question

Locus of centres of the circles which touche the two circles x2+y2=16 and x2+y2=16x externally is

A
12x2+12y292x+144=0
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B
12x2+4y296x+144=0
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C
x2+y296x+144=0
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D
12x24y296x+144=0
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Solution

The correct option is D 12x24y296x+144=0
Given,
Circle C1:x2+y2=16
Center(0,0) and radius=4

Circle C2:x2+y2=16x
center (8,0) and radius=8

Let r be the radius and (h,k) be the center of required circle

r+4=h2+k2 (1)

r+8=(h8)2+k2 (2)

Subtract eqs.(1) from (1)-

4=(h8)2+k2h2+k2

4+h2+k2=(h8)2+k2

Squaring on Both sides

16+h2+k2+8h2+k2=(h8)2+k2

16+h2+8h2+k2=h2+6416h

8h2+k2=4816h

h2+k2=62h
Squaring on both sides
(h2+k2)=36+4h224h3h2k224h+36=0

Therefore the locus is
4(3x2y224x+36=0)
12x24y296x+144=0

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