Question

Find the center and the radius of the circle with the given equation.

Problem 1) x^2+y^2+9x-8y+4=0
Problem 2) x^2+y^2-6y=0

Answer 1

ok we want to put the circles into standard form:
(x - a)^2 + (y - b)^2 = r^2
where the center is (a,b) and the radius is r

so for #1:
x^2 + y^2 + 9x - 8y + 4 = 0
x^2 + 9x + y^2 - 8y = 4

we complete the square for x,y:
x^2 + 9x + 81/4 + y^2 - 8y + 16 = 4 + 81/4 + 16 = 161/4
(x + 9/2)^2 + (y - 4)^2 = 161/4
so (a,b) = (9/2, 4) and r = sqrt(161/4) = sqrt(161)/2

for #2:
x^2+y^2-6y=0
x^2 + y^2 - 6y + 9 = 9
x^2 + (y - 3)^2 = 9
so (a,b) = (0,3) and r = sqrt(9) = 3

Answer 2

err for #1 i meant x^2 + 9x + y^2 - 8y = -4 so r^2 = -4 + 81/4 + 16 = 81/4 + 12 = 129/4
meaning r = sqrt(129)/2