# Complete the equation?

• Dec 2nd 2012, 01:21 PM
M670
Complete the equation?
Complete the equation of the circle centered at http://webwork.mathstat.concordia.ca...81e4e083b1.png that passes through http://webwork.mathstat.concordia.ca...73d234dcf1.png

__________=0

I know I need to get into the form $x^2+Dx+y^2+Ey+F=R^2$ just not sure ?
I am at this point $(x+7)^2+(y-9)^2=r^2$ I need help to expand it out
• Dec 2nd 2012, 01:46 PM
Plato
Re: Complete the equation?
Quote:

Originally Posted by M670
Complete the equation of the circle centered at http://webwork.mathstat.concordia.ca...81e4e083b1.png that passes through http://webwork.mathstat.concordia.ca...73d234dcf1.png __________=0
I know I need to get into the form $x^2+Dx+y^2+Ey+F=R^2$ just not sure ? I am at this point $(x+7)^2+(y-9)^2=r^2$ I need help to expand it out

YES and $r^2=(-7-4)^2+(9-1)^2$

You should know the basic algebra: $(x+7)^2=x^2+14x+49$.
• Dec 2nd 2012, 01:48 PM
Soroban
Re: Complete the equation?
Hello, M670!

Quote:

$\text{Find the equation of the circle centered at }C(\text{-}7,9)\text{ that passes through }P(4,1).$

You're making hard work out of it . . .

The equation of a circle with center $(h,k)$ and radius $r$ is: . $(x-h)^2 + (y-k)^2 \:=\:r^2$

You already know the center: . $C(\text{-}7,9)$

You need only the radius . . . $r \:=\:\overline{CP}$
. . Use the Distance Formula.
• Dec 2nd 2012, 01:58 PM
M670
Re: Complete the equation?
Quote:

Originally Posted by Plato
YES and $r^2=(-7-4)^2+(9-1)^2$

You should know the basic algebra: $(x+7)^2=x^2+14x+49$.

So I came up with $x^2+14x+y^2-18y+130=0$ but that still isn't right and I dont know why? I need to write formula =0
• Dec 2nd 2012, 02:05 PM
Plato
Re: Complete the equation?
Quote:

Originally Posted by M670
So I came up with $x^2+14x+y^2-18y+130=0$ but that still isn't right and I dont know why? I need to write formula =0

You did not use the fact $r^2=185$, did you?
• Dec 2nd 2012, 03:02 PM
M670
Re: Complete the equation?
Quote:

Originally Posted by Plato
You did not use the fact $r^2=185$, did you?

No I did not use the $r^2=185$ as I know the circle formula to be [tex](x-h)^2+(y-K)^2=r^2[/tex} since it's asking me for the equation equal to zero I am not sure where I would put radius?
• Dec 2nd 2012, 03:13 PM
Plato
Re: Complete the equation?
Quote:

Originally Posted by M670
No I did not use the $r^2=185$ as I know the circle formula to be [tex](x-h)^2+(y-K)^2=r^2[/tex} since it's asking me for the equation equal to zero I am not sure where I would put radius?

I don't think any of us can help you understand this question.
Please sit down with a live tutor. Go over what this question means.
• Dec 2nd 2012, 03:17 PM
M670
Re: Complete the equation?
Quote:

Originally Posted by Plato
I don't think any of us can help you understand this question.
Please sit down with a live tutor. Go over what this question means.

Thanks for trying, Its this web program called webworks, It really sucks as you have to answer it a specific way or esle it wont except it.
• Dec 2nd 2012, 03:19 PM
Prove It
Re: Complete the equation?
You know that the centre is \displaystyle \begin{align*} (-7, 9) \end{align*} and one point on your circle is \displaystyle \begin{align*} (4, 1) \end{align*}. The radius is the distance between the centre to any point on the circle, so how can you work out the distance between these two points?
• Dec 2nd 2012, 03:51 PM
M670
Re: Complete the equation?
Quote:

Originally Posted by Prove It
You know that the centre is \displaystyle \begin{align*} (-7, 9) \end{align*} and one point on your circle is \displaystyle \begin{align*} (4, 1) \end{align*}. The radius is the distance between the centre to any point on the circle, so how can you work out the distance between these two points?

Yes I know I can use the standard distance formula which will give my radius of the circle but the question was asking me the equation of a circle with its center at (-7,9) and passes throught points (4,1)
• Dec 2nd 2012, 04:07 PM
Plato
Re: Complete the equation?
Quote:

Originally Posted by M670
Yes I know I can use the standard distance formula which will give my radius of the circle but the question was asking me the equation of a circle with its center at (-7,9) and passes throught points (4,1)

Well damn it, use it!
• Dec 2nd 2012, 04:08 PM
Prove It
Re: Complete the equation?
I don't think you've bothered reading the posts above you. If you have the centre and radius of your circle, substitute them into this:

\displaystyle \begin{align*} (x - h)^2 + (y -k)^2 = r^2 \end{align*}

where \displaystyle \begin{align*} (h, k) \end{align*} is the centre of your circle and \displaystyle \begin{align*} r \end{align*} is the radius.

This is the general form of the equation of a circle. But since you are asked to set the equation equal to 0, you expand everything and move everything to one side.
• Dec 2nd 2012, 04:35 PM
M670
Re: Complete the equation?
Ok So $h=-7 k=9 r= \sqrt185$ $(x+7)^2+(y-9)^2=\sqrt185$ which is then $x^2+14x+49+y^2-18y+81=\sqrt185$

HOLY SH** it just dawned on me $x^2+14x+y^2-18y-55=0$ is the correct answer....
Let's hope it doesn't take me this long on my final...
Many thanks Prove it and Plato...
• Dec 2nd 2012, 05:56 PM
Prove It
Re: Complete the equation?
Actually \displaystyle \begin{align*} r = \sqrt{185} \end{align*} so \displaystyle \begin{align*} r^2 = 185 \end{align*} which means \displaystyle \begin{align*} (x + 7)^2 + (y - 9)^2 = 185 \end{align*}.