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The locus of the centres of the circles,...

The locus of the centres of the circles, which touch the circle, ` x^(2)+y^(2)=1` externally, also touch the Y-axis and lie in the first quadrant, is

A

`y=sqrt(1+2x,)xge0`

B

`y=sqrt(1+4x,)xge0`

C

`y=sqrt(1+2y,)yge0`

D

`y=sqrt(1+4y,)yge0`

Text Solution

Verified by Experts

The correct Answer is:
A
Let (h,k) be the centre of the circle and radius r = h as circle touch the Y-axis and other circle `x^(2)+y^(2) = 1` whose centre (0,0) and radius is 1.

`therefore OC = r + 1`
[`therefore` if circles touch each other externally, then `C_(1)C_(2)=r_(1)+r_(2)`]
`rArr sqrt(h^(2)+k^(2))=h+1, hgt0`
and `kgt0`, for first quadrant.
`rArr h^(2)+k^(2)=h^(2)+2h+1`
`rArr k^(2)=2h+1`
`rArrk=sqrt(1+2h), as kgt0`
Now, on taking locus of centre (h,k),we get
`y = sqrt(1+2x),xge0`
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