JEE Advance - Mathematics (1980 - No. 22)
Two circles $${x^2} + {y^2} = 6$$ and $${x^2} + {y^2} - 6x + 8 = 0$$ are given. Then the equation of the circle through their points of intersection and the point (1, 1) is
$${x^2} + {y^2} - 6x + 4 = 0$$
$${x^2} + {y^2} - 3x + 1 = 0$$
$${x^2} + {y^2} - 4y + 2 = 0$$
none of these