A few days ago
jamie68117

How would I find the radius of the circle with equation x^2 +y^2 +2x-8y-4=0?

How would I find the radius of the circle with equation x^2 +y^2 +2x-8y-4=0?

I’m pretty sure I have to complete the square, but I’m not really sure how to do that. I think to complete the square of the x I need to add a 1 to both sides, but what do I need to do to complete the square of the y?!

If you answer, please explain your answer.

THANKS!

Top 3 Answers
A few days ago
wayner122

Favorite Answer

Yes you are going to have to complete the square which is sort of tough to explain in writing but here goes nothing:

First you want to group the x and y terms together and move the constant term to the other side:

(x^2 +2x) + (y^2 – 8y) = 4

Now you have to complete the square for x and for y:

What you do is take the term that is not squared in each group: So first look at the 2x. Take the 2 and divide it by 2 (2/2 = 1 and square it. 1^2 = 1. Add this number to your bracket and to the other side. So you will have (x^2 + 2x +1) + (y^2 – 8y) = 4 + 1. Now you want to do the same thing in your other group. So look at the -8y term. You take half of -8 which is -4 and square it to get 16. So you add 16 to both sides. (x^2 +2x +1) + (y^2 -8y +16) = 4 + 1 + 16. Now you have perfect squares on the left hand side. So factor the groups to get (x + 1)^2 + (y – 4)^2 = 21. From this you can tell the center of the circle and the radius. The radius is the square root of the number on the right hand side. r = sqroot(21).

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A few days ago
Anonymous
forget the nubers use pi r squared
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A few days ago
Anonymous
what is wrong with pi R squared?

answer though – you dont.

WHAT ARE YOU DOING WITH ALL THOSE NUMBERS!!!! YOUVE STOLEN THEM – someone might need them back!

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