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If two circles (x+4)^(2)+y^(2)=1 and (x-...

If two circles `(x+4)^(2)+y^(2)=1 and (x-4)^(2)+y^(2)=9` are touched extermally by a circle, then locus of centre of variable circle is

A

`(x^(2))/(15)-(y^(2))/(1)=1`

B

`(x^(2))/(4)-(y^(2))/(12)=1`

C

`(x^(2))/(1)-(y^(2))/(15)=1`

D

`(x^(2))/(12)-(y^(2))/(4)=1`

Text Solution

Verified by Experts

The correct Answer is:
C

`CS=r+1`
`CS'=r+3`
`CS'-CS=2`
Locus of C is hyperbola with `(-4,0)` and `(4,0)` as foci,
Here, 2a=2, so a = 1.
Also, 2ae = 8, so ae = 4
`therefore" "b^(2)=a^(2)(e^(2)-1)=15`
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