Center and Radius of Circles with Given Equations

[cheab14]

The question states:
Find the centre and radius of each circle with equations as given
(a) 3x^2 + 3y^2 = 81
(b) x^2 = 6y - y^2

I really don't know how to approach this question, i started (a) by dividing both sides by 3 but then i don't know where to go from there, and i don't even know how to do (b)...please help me

 

[cristo]

Well, you know that a circle with centre (p,q) and radius r can be written (x-p)²+(y-q)²=r².

(a) once you've divided through by 3, it is in the above form, isn't it?
(b) try to put it in the above form by, maybe, completing the square for the y component.

 

[cheab14]

ok but cristo, for (a) you're saying that my centre would just be p and q?..no actual numbers?...and for (b) my radius would be 9? -also for (b) what would be my centre after completing the square for the y component?

 

[Feldoh]

a) P and Q are dummy variables. In your first problem: 3x^2 + 3y^2 = 81 you could think of this as 3(x-0)^2 + 3(y-0)^2 = 81

b) Remember the equation is (x-p)^2+(y-q)^2=r^2, make sure you're accounting for the radius^2.

The center is straight forward once you have completed the square.

 

[cheab14]

...thank u guys