Question: Write
in standard form.
Answer:
![(x-0)^2+(y+1)^2=36](https://tex.z-dn.net/?f=%28x-0%29%5E2%2B%28y%2B1%29%5E2%3D36)
Step-by-step explanation:
You group all your terms that have a
in it together.
You group all your terms that have a
in it together.
You put everything that doesn't have a variable on the opposing side.
Let's begin.
![x^2+y^2+2y-35=0](https://tex.z-dn.net/?f=x%5E2%2By%5E2%2B2y-35%3D0)
Add 35 on both sides:
![x^2+y^2+2y=35](https://tex.z-dn.net/?f=x%5E2%2By%5E2%2B2y%3D35)
Group terms together that have a
in it and do the same for the
terms.
![(x^2)+(y^2+2y)=35](https://tex.z-dn.net/?f=%28x%5E2%29%2B%28y%5E2%2B2y%29%3D35)
The neat thing here is we don't really need to do anything for the
's since ![x^2=(x)^2=(x-0)^2](https://tex.z-dn.net/?f=x%5E2%3D%28x%29%5E2%3D%28x-0%29%5E2)
![(x-0)^2+(y^2+2y+?)=35+?](https://tex.z-dn.net/?f=%28x-0%29%5E2%2B%28y%5E2%2B2y%2B%3F%29%3D35%2B%3F)
Whatever we add to one side we must add to the other.
[Side note: You are going to complete the squares for the
terms using the following identity:
.
]
So we are going to add
on both sides (notice the coefficient of
is 2 and the coefficient of
is 1).
![(x-0)^2+(y^2+2y+(\frac{2}{2})^2)=35+(\frac{2}{2})^2](https://tex.z-dn.net/?f=%28x-0%29%5E2%2B%28y%5E2%2B2y%2B%28%5Cfrac%7B2%7D%7B2%7D%29%5E2%29%3D35%2B%28%5Cfrac%7B2%7D%7B2%7D%29%5E2)
Applying the right hand side of the identity in my side note:
![(x-0)^2+(y+\frac{2}{2}))^2=35+(1)^2](https://tex.z-dn.net/?f=%28x-0%29%5E2%2B%28y%2B%5Cfrac%7B2%7D%7B2%7D%29%29%5E2%3D35%2B%281%29%5E2)
![(x-0)^2+(y+1)^2=35+1](https://tex.z-dn.net/?f=%28x-0%29%5E2%2B%28y%2B1%29%5E2%3D35%2B1)
![(x-0)^2+(y+1)^2=36](https://tex.z-dn.net/?f=%28x-0%29%5E2%2B%28y%2B1%29%5E2%3D36)
The cool thing about standard form is that it tells you the center and the radius.
In this case the center is (0,-1) and the radius is 6.
Standard form of a circle is
where
is the center and
is the radius.