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.
Best Answer
Your book is right. Consider a circumference of radius 4 centered at $(0,0)$. Its equation is:
$$ x^2 + y^2 = 4^2 = 16 $$
Now, with your circumference centered at $(0,5)$ we get:
$$ x^2 + (y-5)^2 = 16 $$
$$ x^2 + y^2 - 10y + 25 - 16 = 0 $$
$$ x^2 + y^2 - 10y + 9 = 0 $$