Equation of a circle help. (1 Viewer)

[FTW]

Hey, I was just study for an assessment and I am faced with the question for the equation of a circle. I missed the lesson in class and I know the general idea but I am stuck on this question:


x (squ) - 6x + y (squ) + 12y + 36 = 0


If anyone could help me with this and show the working, that would be great! Thanks in advance.

 

[annabackwards]

x^2 - 6x + y^2 + 12y + 36 = 0
x^2 - 6x + y^2 + 12y = -36
x^2 - 6x + 9 + y^2 + 12y + 36 = -36 + 9 + 36
(x^2 - 6x + 9) + (y^2 + 12y + 36) = 9
(x - 3)^2 + (y + 6)^2 = 9

You're basically completing the square for x^2 - 6x and y^2 + 12y; just remember that if you add on the LHS you have to add on the RHS

 

[muzeikchun852]

Hey, I was just study for an assessment and I am faced with the question for the equation of a circle. I missed the lesson in class and I know the general idea but I am stuck on this question:


x (squ) - 6x + y (squ) + 12y + 36 = 0


If anyone could help me with this and show the working, that would be great! Thanks in advance.

x^2 - 6x + 9 - 9 + y^2 + 12y + 36 - 36 + 36 = 0
(x - 3)^2 + (y + 6)^2 - 9 - 36 + 36 = 0
(x - 3)^2 + (y + 6)^2 = 9

centre = 3,6
radius = 3

 

[Top Secret]

x^2 - 6x + 9 - 9 + y^2 + 12y + 36 - 36 + 36 = 0
(x - 3)^2 + (y + 6)^2 - 9 - 36 + 36 = 0
(x - 3)^2 + (y + 6)^2 = 9

centre = 3,6
radius = 3

Centre: (3,-6)

 

[FTW]

tyvm