Complete the equation?

**M670** · December 2nd 2012, 01:21 PM

Complete the equation of the circle centered at

that passes through

__________=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

**Plato** · December 2nd 2012, 01:46 PM

Originally Posted by M670

Complete the equation of the circle centered at

that passes through

__________=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: $(x+7)^2=x^2+14x+49$ .

**Soroban** · December 2nd 2012, 01:48 PM

Hello, M670!

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

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.

**M670** · December 2nd 2012, 01:58 PM

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

**Plato** · December 2nd 2012, 02:05 PM

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?

**M670** · December 2nd 2012, 03:02 PM

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?

**Plato** · December 2nd 2012, 03:13 PM

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.
Your misunderstanding is very profound.
Please sit down with a live tutor. Go over what this question means.

**M670** · December 2nd 2012, 03:17 PM

Originally Posted by Plato

I don't think any of us can help you understand this question.
Your misunderstanding is very profound.
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.

**Prove It** · December 2nd 2012, 03:19 PM

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?

**M670** · December 2nd 2012, 03:51 PM

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)

**Plato** · December 2nd 2012, 04:07 PM

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!

**Prove It** · December 2nd 2012, 04:08 PM

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.

**M670** · December 2nd 2012, 04:35 PM

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...

**Prove It** · December 2nd 2012, 05:56 PM

Actually $\displaystyle \begin{align*} r = \sqrt{185} \end{align*}$ so $\displaystyle \begin{align*} r^2 = 185 \end{align*}$ which means $\displaystyle \begin{align*} (x + 7)^2 + (y - 9)^2 = 185 \end{align*}$ .

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