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The centre of family of circles cutting ...

The centre of family of circles cutting the family of circles `x^2 + y^2 + 4x(lambda-3/2) + 3y(lambda-4/3)-6(lambda+2)=0` orthogonally, lies on

A

`x - y - 1 = 0`

B

`4x +3y - 6 = 0`

C

`4x +3y +7 = 0`

D

`3x - 4y - 1 = 0`

Text Solution

Verified by Experts

The correct Answer is:
B


`(x^(2) +y^(2) -6x -4y -12) + lambda (4x +3y -6) =0`
This is family of circles passing through points of intersection of circle `x^(2) + y^(2) -6x -4y -12 =0` and line `4x +3y -6 =0`. Other family will cut this family at A and B.
Hence, locus of centre of circle of other family is this common chord `4x +3y -6 =0`.
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