The centre of the circle must lie on the perpendicular bisector of PQ. This is the line through the midpoint (3,–1) of PQ, with gradient 2. So its equation is .
If a circle with centre at the point touches the y-axis then its radius must be equal to x. If that circle passes through the point (1,0) then (the square of the distance from to (1,0)) must therefore be equal to .
Solve the equation to find the x-coordinate of the centre of the circle (two solutions, because it is a quadratic equation).