SOLUTION: the equation is: 2x^2+2y^2+4y=0 and I am supposed to find the center and the radius of the circle as well as the x and y intercepts but I am not sure how to find the center and rad
Algebra.Com
Question 783437: the equation is: 2x^2+2y^2+4y=0 and I am supposed to find the center and the radius of the circle as well as the x and y intercepts but I am not sure how to find the center and radius. If you could help me find those I can find the x and y intercepts. Thanks
Found 2 solutions by stanbon, solver91311:
Answer by stanbon(75887) (Show Source): You can put this solution on YOUR website!
2x^2+2y^2+4y=0 and I am supposed to find the center and the radius of the circle
-----------------
Complete the square on the x-terms and on the y-terms:::
2(x^2) + 2(y^2+2y) = 0
-----
2(x-0)^2 + 2(y^2+2y+1) = 2
------
2(x-0)^2 + 2(y+1)^2 = 2
-----
(x-0)^2 + (y+1)^2 = 1
-----
Center: (0,-1)
Radus : sqrt(2)
==================
Cheers,
Stan H.
==================
Answer by solver91311(24713) (Show Source): You can put this solution on YOUR website!
The equation of a circle with center at )
and radius 
is ^2\ +\ \left(y\ -\ k\right)^2\ =\ r^2)
. So to address your situation, you need to make 
look like . Actually, much easier than it looks.
The process requires you to complete the square on each of the two variables. For this particular problem, dealing with 
is trivial since there is no first degree term. We know from that fact that the value of 
, the -coordinate of the center of the circle, is 
.
The first step in the process is to divide by the lead coefficient -- the one on the 
term. Of course, if this is to be a circle, the coefficient on the 
term has to be the same. So you end up with:
Next, divide the coefficient on the 1st degree 
term by 2 and square the result. Add that result to both sides of the equation. 2 divided by 2 is 1. 1 squared is 1, so:
Now the 
term is a perfect square and the 
terms and the constant in the LHS form a perfect square, so factor:
^2\ =\ 1)
Next, re-write everything so that it fits the circle equation pattern:
^2\ +\ \left(y\ -\ (-1)\right)^2\ =\ 1^2)
Now you can determine the center and radius by inspection:
Center: )
Radius: 
John
Egw to Beta kai to Sigma
My calculator said it, I believe it, that settles it
RELATED QUESTIONS
x^2 + y^2 -2x +2y -18 = 0
I need to find the center and radius of the circle (answered by josgarithmetic)
I am confused as to how to find the radius and center of a circle by using an equation.... (answered by rapaljer)
find the center and radius of the circle having equation... (answered by Edwin McCravy)
given the equation of a circle x^2+y^2-10x+4y+13=0. find its center and radius. i am... (answered by Theo)
find the center and radius of the circle with the equation... (answered by solver91311)
i am supposed to write an equation of a circle with the center at the origin and radius... (answered by Alan3354)
Find the center and radius of the circle... (answered by solver91311)
Find the center and radius of the circle
{{{... (answered by richwmiller)
Find the center and radius of the circle... (answered by ewatrrr)
