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Question

If the circles x2+y2−2x−4y=0 and x2+y2−8y−k=0 touches each other internally, then the possible value of k is

A
5
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B
4
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C
3
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D
2
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Solution

The correct option is B 4
Given circles are
S1:x2+y22x4y=0 and S2:x2+y28yk=0
Now, the centre and radius of the circles are
C1=(1,2), r1=5C2=(0,4), r2=16+k
For the square root to be defined,
16+k0k16
Now,
C1C2=12+22=5
As both the circles touches internally, so
C1C2=|r1r2|5=|516+k|516+k=±516+k=5±516+k=25,0k=16,4

Hence, from the given options k=4.

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