Question

How do you write the equation in standard form for the circle x² + y² + 2y - 35 = 0

Answer 1

Question: Write $x^2+y^2+2y-35=0$ in standard form.

Answer:

$(x-0)^2+(y+1)^2=36$

Step-by-step explanation:

You group all your terms that have a $x$ in it together.

You group all your terms that have a $y$ in it together.

You put everything that doesn't have a variable on the opposing side.

Let's begin.

$x^2+y^2+2y-35=0$

Add 35 on both sides:

$x^2+y^2+2y=35$

Group terms together that have a $x$ in it and do the same for the $y$ terms.

$(x^2)+(y^2+2y)=35$

The neat thing here is we don't really need to do anything for the $x$'s since $x^2=(x)^2=(x-0)^2$

$(x-0)^2+(y^2+2y+?)=35+?$

Whatever we add to one side we must add to the other.

[Side note: You are going to complete the squares for the $y$ terms using the following identity:

$u^2+bu+(\frac{b}{2})^2=(u+\frac{b}{2})^2$.

]

So we are going to add $(\frac{2}{2})^2$ on both sides (notice the coefficient of $y$ is 2 and the coefficient of $y^2$ is 1).

$(x-0)^2+(y^2+2y+(\frac{2}{2})^2)=35+(\frac{2}{2})^2$

Applying the right hand side of the identity in my side note:

$(x-0)^2+(y+\frac{2}{2}))^2=35+(1)^2$

$(x-0)^2+(y+1)^2=35+1$

$(x-0)^2+(y+1)^2=36$

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 $(x-h)^2+(y-k)^2=r^2$ where $(h,k)$ is the center and $r$ is the radius.