# Form a Circle Equation

• October 7th 2008, 11:03 AM
WhoaBlackBetty
Form a Circle Equation
Could someone please double check this for me and see if I am doing it correctly.
Here is what the equation is and I'm supposed to find is the center point for (x,y) and the radius of the circle.

2x^2 + 2y^2 + 20x +16y + 10 =0

2x^2 + 2y^2 +20x + 16y = -10

Then

2x^2/2 + 2y^2/2 +20x/2 + 16y/2 = -10/2

Then

x^2 + y^2 + 10x + 8y = -5

Then

(x^2+10x+25) + (y^2 + 8y + 16) = -5 + 25 + 16

Then

(x+5)^2 + (y+4)^2 = 36^2

Which yields my answer of ....

(5,4) with a radius of 6.

Please let me know if I am doing these correctly... I have no way to check my answer thus no way to know if i am on the right path.
• October 7th 2008, 11:08 AM
earboth
Quote:

Originally Posted by WhoaBlackBetty
Could someone please double check this for me and see if I am doing it correctly.
Here is what the equation is and I'm supposed to find is the center point for (x,y) and the radius of the circle.

2x^2 + 2y^2 + 20x +16y + 10 =0

2x^2 + 2y^2 +20x + 16y = -10

Then

2x^2/2 + 2y^2/2 +20x/2 + 16y/2 = -10/2

Then

x^2 + y^2 + 10x + 8y = -5

Then

(x^2+10x+25) + (y^2 + 8y + 16) = -5 + 25 + 16

Then

(x+5)^2 + (y+4)^2 = 36 = 6^2

Which yields my answer of ....

(-5, -4) with a radius of 6. Don't miss the negative signs!

Please let me know if I am doing these correctly... I have no way to check my answer thus no way to know if i am on the right path.

...
• October 7th 2008, 12:20 PM
WhoaBlackBetty
Thanks, I was close. How does that turn negative? I think thats why I screwed up. Please elaborate on that... I'm not getting when it is positive or negative.
• October 7th 2008, 10:41 PM
earboth
Quote:

Originally Posted by WhoaBlackBetty
Thanks, I was close. How does that turn negative? I think thats why I screwed up. Please elaborate on that... I'm not getting when it is positive or negative.

A circle is defined as the locus of points which have a constant distance to a given fixed point.

Let be the fixed point M(h, k) (it's the midpoint of the circle) and the constant distance r (it's the radius of the circle). Now use the distance formula to calculate the distance between any point on the circle P(x, y) and the midpoint M:

$r^2 = (x-h)^2+(y-k)^2$

You see that the operation sign in the brackets must be always minus.

In your question you have got:

$(x+5)^2 + (y+4)^2 = 6^2~\implies~(x-(-5))^2+(y-(-4))^2=6^2$ and now you can see that the point M is M(-5, -4).
• October 8th 2008, 03:39 AM
WhoaBlackBetty
Ok...Thanks a lot. So, it is since the original circle formula is negative it needs the other negative to be positive. So, whatever I pullout will be the opposite of it's sign. Thanks! That helps out a lot.