Asked by Sam
Which is the standard form of a circle with equation x^2 + 4x + y^2 - 6y = 3?
A. (x+2)^2+(y+9)^2=16
B. (x-2)^2+(y-9)^2=3
C. (x+2)^2+(y-3)^=16
D. (x+2)^2+(y-3)^2=3
A. (x+2)^2+(y+9)^2=16
B. (x-2)^2+(y-9)^2=3
C. (x+2)^2+(y-3)^=16
D. (x+2)^2+(y-3)^2=3
Answers
Answered by
Steve
Nope. Forgot to add the constants to both sides of the equation.
x^2 + 4x + y^2 - 6y = 3
(x^2+4x+4) + (y^2-3y+9) = 3+4+9
(x+2)^2 + (y-3)^2 = 16
(C)
x^2 + 4x + y^2 - 6y = 3
(x^2+4x+4) + (y^2-3y+9) = 3+4+9
(x+2)^2 + (y-3)^2 = 16
(C)
There are no AI answers yet. The ability to request AI answers is coming soon!
Submit Your Answer
We prioritize human answers over AI answers.
If you are human, and you can answer this question, please submit your answer.