A circle has to points on the circumference (0, 1) and (0, 9) that bisect the circle and I have to give the equation of the circle in general form.

So I can find the midpoint of the circle as ${(0, {{9 + 1} \over 2})}$ = (0, 5).

Using the distance formula to find the distance between the two points I get ${\sqrt {(0)^2 + (9-1)^2}}$ which is 8, divided by 2 is 4.

So I would give the equation as:

${x^2 + y^2 -10y + 16 = 0}$

But the book says

${x^2 + y^2 -10y +9 = 0}$

I don't get how the radius is 3.