# Standard form & Graphing

• Jan 25th 2009, 05:32 AM
AlgebraicallyChallenged
Standard form & Graphing
Good morning forum I need some assistance with the following question:

Put the equation x2 + y2 + 8x + 2y = 29 into standard form and sketch the graph of the equation.

Thanks AC
• Jan 25th 2009, 05:57 AM
Prove It
Quote:

Originally Posted by AlgebraicallyChallenged
Good morning forum I need some assistance with the following question:

Put the equation x2 + y2 + 8x + 2y = 29 into standard form and sketch the graph of the equation.

Thanks AC

I take it that the 2's are superscript.

So it should read $x^2 + y^2 + 8x + 2y = 29$.

First off, this is the equation of a circle.

The standard form of a circle is $(x - h)^2 + (y - k)^2 = r^2$ where h is the x-coordinate of the centre, k is the y-coordinate of the centre, and r is the radius.

To get it into this standard form, complete the square on the x terms and the y terms.

$x^2 + 8x + y^2 + 2y = 29$

$x^2 + 8x + 4^2 + y^2 + 2y + 1^2 = 29 + 4^2 + 1^2$

$(x + 4)^2 + (y + 1)^2 = 46$

$[x - (-4)]^2 + [y - (-1)]^2 = (\sqrt{46})^2$.

So what's the centre? What's the radius? Can you sketch the circle?
• Jan 25th 2009, 06:04 AM
andreas
Quote:

Originally Posted by AlgebraicallyChallenged
Good morning forum I need some assistance with the following question:

Put the equation x2 + y2 + 8x + 2y = 29 into standard form and sketch the graph of the equation.

Thanks AC

You have to complete squares, 8th grade trick.

$x^2+8x=(x+4)^2-16$ and $y^2+2y=(y+1)^2-1$

substituting this and rearranging the following is obtained:

$(x+2)^2+(y+1)^2=29+16+1=46$

this is a circle with center at $(-2,-1)$ and radius $\sqrt{46}$
• Jan 25th 2009, 06:20 AM
Prove It
Quote:

Originally Posted by andreas
You have to complete squares, 8th grade trick.

$x^2+8x=(x+4)^2-16$ and $y^2+2y=(y+1)^2-1$

substituting this and rearranging the following is obtained:

$(x+2)^2+(y+1)^2=29+16+1=46$

this is a circle with center at $(-2,-1)$ and radius $\sqrt{46}$

Wrong, it's a circle with centre at $(-4, -1)$ and radius $\sqrt{46}$.
• Jan 25th 2009, 07:49 AM
andreas
Quote:

Originally Posted by Prove It
Wrong, it's a circle with centre at $(-4, -1)$ and radius $\sqrt{46}$.

Yes I admit, it was typo...