# sketching a graph

• October 4th 2009, 03:30 PM
Candy101
sketching a graph
how can i sketch x^2+y^2-16x+8y+16=0

can you give me please try to explain the steps that i need to do

thanks
• October 4th 2009, 03:47 PM
skeeter
Quote:

Originally Posted by Candy101
how can i sketch x^2+y^2-16x+8y+16=0

can you give me please try to explain the steps that i need to do

thanks

it's a circle ... complete the square for both x and y and get it into the form

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

(h,k) is the center ... r is the radius
• October 4th 2009, 03:47 PM
artvandalay11
Quote:

Originally Posted by Candy101
how can i sketch x^2+y^2-16x+8y+16=0

can you give me please try to explain the steps that i need to do

thanks

this is a circle that needs to be put in standard form to easily sketch it. I was able to realize that since the coefficients of the $x^2$ and $y^2$ are the same, namely =1

We must complete the square for both x and y, so we divide the coefficient of the x and y terms by 2, square them, and add them to both sides

$x^2+y^2-16x+8y+16=x^2-16x+(-8)^2+y^2+8y+4^2+16=8^2+4^2$

And now we can factor

$(x-8)^2+(y+4)^2=64+16-16=64$

Can you sketch that now or do you need further assistance?
• October 4th 2009, 04:28 PM
Candy101
i dont really know how to sketch it...
• October 4th 2009, 06:17 PM
pacman
http://www.mathhelpforum.com/math-he...f6c2340a-1.gif

(x - 8)^2 + (y + 4)^2 = 64 = 8^2

compare it with the standard equation of a circle, that is

(x - h)^2 + (y + 4)^2 = r^2,

h,k is the CORDINATE of the center of the CIRCLE, where r is the radius.

from (x - 8)^2 + (y + 4)^2 = 64 = 8^2; center coordinate is (h,k) is (8,-4) and r = 8.

see the graph . . . .

(Giggle)

if the standard equation of the circle awe you, resort to point plotting, maybe you have not encounter yet ANALYTIC GEOMETRY.

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