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mixas84 [53]
3 years ago
5

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

Mathematics
1 answer:
Ierofanga [76]3 years ago
4 0

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.

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