JEE Advance - Mathematics (1993 - No. 17)
The locus of the centre of a circle, which touches externally the circle $${x^2} + {y^2} - 6x - 6y + 14 = 0$$ and also touches the y-axis, is given by the equation:
$${x^2} - 6x - 10y + 14 = 0$$
$${x^2} - 10x - 6y + 14 = 0$$
$${y^2} - 6x - 10y + 14 = 0$$
$${y^2} - 10x - 6y + 14 = 0$$