write the equation x^2-2x+y^2+4y=11 in standard form?

4 ⤊

a. (x-1)^2+(y+2)^2=16
b. (x+1)^2+(y-2)^2=16
c. (x+1)^2 /1 + (y-2)^2 /4=1
d. (x-1)^2 /4 + (y-1)^2 /4=1

2007-01-31 03:30:00 · 5 answers · asked by Anna 1 in Science & Mathematics ➔ Mathematics

x^2-2x+y^2+4y=11
x^2-2x+1+y^2+4y+4=11+1+4
(x-1)^2 + (y+2)^2 = 16
(answer is a)

2007-01-31 03:37:23 · answer #1 · answered by seah 7 · 3⤊ 0⤋

xÂ² - 2x + yÂ² + 4y = 11
(xÂ² - 2x + 1) + (yÂ² + 4y + 4) = 11 + 1 + 4
(x - 1)Â² + (y + 2)Â² = 16
Answer a)

2007-01-31 12:08:48 · answer #2 · answered by Como 7 · 0⤊ 0⤋

x^2-2x+1+y^2+4y+4=16
so a

2007-01-31 11:45:38 · answer #3 · answered by Tamino 2 · 0⤊ 0⤋

The standard form is a) .It represents a circunference with center (1,-2) and radius 4

2007-01-31 11:56:10 · answer #4 · answered by santmann2002 7 · 0⤊ 0⤋

2007-01-31 11:36:44 · answer #5 · answered by hratanasov 1 · 1⤊ 0⤋