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Two circle of equal radii are intersect ...

Two circle of equal radii are intersect at `(0, 1)` and `(0, ?1)`, and the tangent at `(0, 1)` to one of those passes through the centre of other. Then the distance between the centres is equal to (A) 2` (B) `sqrt2` (C) `1/sqrt2` (D) `1/2`

A

`sqrt2`

B

`2sqrt2`

C

1

D

2

Text Solution

Verified by Experts

The correct Answer is:
D
Clearly, circles are orthogonal because tangent at one point of intersection is passing through centre of the other.

Let `C_(1)(alpha, 0) and C_(2)(-alpha, 0)` are the centres.
then, `S_(1)-=(x-alpha)^(2)+y^(2)=alpha^(2)+1`
`rArr S_(1)-=x^(2) + y^(2)-2ax-1 = 0`
`[therefore " radius", r = sqrt((a-0)^(2)+(0-1)^(2))]`
and `S_(2)-=(x+alpha)^(2)+y^(2)=alpha^(2)+1`
`rArr S_(2)-=x^(2)+y^(2)+2ax-1=0`
Now, `2(alpha)(-alpha)+2*0*0=(-1)+(-1)rArralpha=pm1`
`[therefore "condition of orthogonality is " 2g_(1)g_(2)+2f_(1)f_(2)=c_(1)+c_(2)]`
`therefore C_(1)(1,0) and C_(2)(-1,0)rArrC_(1)C_(2)=2`
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