Respuesta :
Question: Write [tex]x^2+y^2+2y-35=0[/tex] in standard form.
Answer:
[tex](x-0)^2+(y+1)^2=36[/tex]
Step-by-step explanation:
You group all your terms that have a [tex]x[/tex] in it together.
You group all your terms that have a [tex]y[/tex] in it together.
You put everything that doesn't have a variable on the opposing side.
Let's begin.
[tex]x^2+y^2+2y-35=0[/tex]
Add 35 on both sides:
[tex]x^2+y^2+2y=35[/tex]
Group terms together that have a [tex]x[/tex] in it and do the same for the [tex]y[/tex] terms.
[tex](x^2)+(y^2+2y)=35[/tex]
The neat thing here is we don't really need to do anything for the [tex]x[/tex]'s since [tex]x^2=(x)^2=(x-0)^2[/tex]
[tex](x-0)^2+(y^2+2y+?)=35+?[/tex]
Whatever we add to one side we must add to the other.
[Side note: You are going to complete the squares for the [tex]y[/tex] terms using the following identity:
[tex]u^2+bu+(\frac{b}{2})^2=(u+\frac{b}{2})^2[/tex].
]
So we are going to add [tex](\frac{2}{2})^2[/tex] on both sides (notice the coefficient of [tex]y[/tex] is 2 and the coefficient of [tex]y^2[/tex] is 1).
[tex](x-0)^2+(y^2+2y+(\frac{2}{2})^2)=35+(\frac{2}{2})^2[/tex]
Applying the right hand side of the identity in my side note:
[tex](x-0)^2+(y+\frac{2}{2}))^2=35+(1)^2[/tex]
[tex](x-0)^2+(y+1)^2=35+1[/tex]
[tex](x-0)^2+(y+1)^2=36[/tex]
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 [tex](x-h)^2+(y-k)^2=r^2[/tex] where [tex](h,k)[/tex] is the center and [tex]r[/tex] is the radius.
