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mollybug16
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
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
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