Question

The equation of a circle in general form is x'+32+6x-4y+4=0.
Write the equation in standard form.
(1 point)
(x+3)3+0v-2)ª=4
(7+3)3+00-2)ª=9
(x+3)'+(2+2)'=4
(7+3)3+00+2)ª=9

Answer 1

Assuming the equation is x^2 + y^2 + 6x - 4y + 4 = 0, complete the square:

(x^2 + 6x) + (y^2 - 4y) + 4 = 0
(x+3)^2 - 9 + (y-2)^2 - 4 + 4 = 0
(x+3)^2 + (y-2)^2 - 9 = 0

So the standard form is (x+3)^2 + (y-2)^2 = 9 (center (-3, 2), radius 3).