Graph the FFG. circle: 4x^2 + 6x + y^2 = 7, y^2 + x^2 + 4x - 4y - 1 = 0, etc.

[SagaSaturn]

I haven't been to school in years. I don't even know the first steps to take. I just want to know the process for one of these.

 

[stapel]

This looks like algebra rather than calculus, so I'm probably misunderstanding things. What is the meaning of "FFG." in the instructions? What are you supposed to be doing with these circle equations?

When you reply, please include a clear listing of your thoughts and efforts so far, so we can see where you're getting stuck. Thank you!

 

[topsquark]

I haven't been to school in years. I don't even know the first steps to take. I just want to know the process for one of these.View attachment 36421

A circle of radius R, with a center (h, k), has the equation
[imath](x-h)^2 + (y-k)^2 = R^2[/imath]

What you need to do is get each of these equations into this format.

So, if you have
[imath]x^2 + 2x + y^2 = 3[/imath]

you need to "complete the square" on [imath]x^2 + 2x[/imath]. Do you recall how to do that?

-Dan

 

[Dr.Peterson]

I haven't been to school in years. I don't even know the first steps to take. I just want to know the process for one of these.View attachment 36421

I suspect that in your dialect, "ffg." is an abbreviation for "following". (It is not in mine.)

As to solving these problems. we typically rewrite the equation (given in what may be called "general form", x^2+y^2+ax+by+c=0) to what may be called "standard form", as shown. (The first, unfortunately, is not a circle because the coefficients of x^2 and y^2 are different; is that a typo?)

Here is one source:

Circle Equations

 

[The Highlander]

I haven't been to school in years. I don't even know the first steps to take. I just want to know the process for one of these.View attachment 36421

The general equation of a circle is (also) given in the form:-

\(\displaystyle x^2+y^2+2gx+2fy+c=0\)   (See here.)​

Where: \(\displaystyle (-g, -f)\) is the centre of the circle

and \(\displaystyle \sqrt{g^2+f^2-c}\)  is its radius.

Have you not been taught this?
The exercise you've been given looks (to me) as if it is based on this 'knowledge'.

I say that because all of the equations you've been given are already in this form
(or very close to it; in 3) & 5) minimal rearrangement is required).

Except, of course, for 1) which is not the equation of a circle!
(I suppose all you can do there is state that.)

So you don't have to get each of the equations into the form: \(\displaystyle (x-h)^2 + (y-k)^2 = R^2\)

Once you have slightly re-arranged 3) & 5) you can (almost) directly "read off " the radius of each circle and the coordinates of its centre which subsequently makes graphing the circles fairly straightforward.

For example:-

Equation 2) is: \(\displaystyle x^{2}+y^{2}+4x-4y-1=0\)

So  \(\displaystyle 2g=4, 2f=-4\) and \(\displaystyle c=-1\\ \implies g=2\) and \(\displaystyle f=-2\).

Therefore, this circle has its centre at (-2, 2) and has radius: \(\displaystyle \sqrt{2^2+ˉ2^2-ˉ1}=\sqrt{4+4+1}=\sqrt{9}=3\) which is then simple enough to graph (as illustrated here).

Now you have been shown the "process" for one, please come back and show us your efforts for the remaining questions: 3) to 5).

Hope that helps. ?

 

[lookagain]

. . .
So  \(\displaystyle 2g=4, 2f=-4\) and \(\displaystyle c=-1\\ \implies g=2\) and \(\displaystyle f=-2\).

Therefore, this circle has its centre at (-2, 2) and has radius: \(\displaystyle \sqrt{2^2+ˉ2^2-ˉ1}=\sqrt{4+4+1}=\sqrt{9}=3\)

\(\displaystyle \sqrt{(2)^2 + (-2)^2 - (-1)} \ = \ \sqrt{4 + 4 + 1} \ = \ \sqrt{9} \ = \ 3 \)

 

[The Highlander]

\(\displaystyle \sqrt{(2)^2 + (-2)^2 - (-1)} \ = \ \sqrt{4 + 4 + 1} \ = \ \sqrt{9} \ = \ 3 \) ???

Are you suggesting that ˉ2 is (somehow) a less valid representation of negative two than (-2)?

 

[lookagain]

Are you suggesting that ˉ2 is (somehow) a less valid representation of negative two than (-2)?

\(\displaystyle -2^2 \ne 4, \ \) but you have included the former in the middle of your radicand.

 

[The Highlander]

\(\displaystyle -2^2 \ne 4, \ \) but you have included the former in the middle of your radicand.

I beg to differ. I did not record \(\displaystyle -2^2 \) anywhere in the radicand!
I recorded ˉ2² (and ˉ2² ≡ ˉ2 × ˉ2 = 4).

 

[pka]

I beg to differ. I did not record \(\displaystyle -2^2 \) anywhere in the radicand!
I recorded ˉ2² (and ˉ2² ≡ ˉ2 × ˉ2 = 4).

I would hope that you are just confused and do not teach the above..
While it is true that [imath](-2)(-2)=(-2)^2=4[/imath], it is false that [imath]-2^2=4.[/imath]
To be fair to @lookagain, you do use [imath]\sqrt{2^2\bf{+^-2^2-^-1}}=\sqrt{4+4+1}[/imath],
when it should be [imath]=\sqrt{4-4+1}[/imath] because [imath]-2^2=-4.[/imath]

 

[The Highlander]

I would hope that you are just confused and do not teach the above..
While it is true that [imath](-2)(-2)=(-2)^2=4[/imath], it is false that [imath]-2^2=4.[/imath]
To be fair to @lookagain, you do use [imath]\sqrt{2^2\bf{+^-2^2-^-1}}=\sqrt{4+4+1}[/imath],
when it should be [imath]=\sqrt{4-4+1}[/imath] because [imath]-2^2=-4.[/imath]

No, I am not at all confused.
I agree wholeheartedly that \(\displaystyle -2^2 = ˉ4\) but I did not write \(\displaystyle -2^2\)!
I quite clearly wrote \(\displaystyle ˉ2^2\) and \(\displaystyle ˉ2\) is just as valid a representation of negative two as \(\displaystyle (-2)\) is! (and, Yes, I do teach that!)
Therefore \(\displaystyle ˉ2^2 = 4\) not \(\displaystyle ˉ4\).
You both appear to be failing to recognize the difference between an operator and a sign!

 

[lookagain]

I, lookagain, am rewriting ˉ2² as ˉ1*2²

<sup>\(\displaystyle \text{equal \ \ to \ \ negative \ \ four.}\)

The exponent operates on the 2 first, and then the result is negated.</sup>

 

[The Highlander]

<sup>\(\displaystyle \text{equal \ \ to \ \ negative \ \ four.}\)

The exponent operates on the 2 first, and then the result is negated.</sup>

Where on earth did I say: "I, lookagain, am rewriting ˉ2² as ˉ1*2²" ???

You are trying to put words in my mouth now?

I have not written any such thing anywhere (because it is utter drivel!)

\(\displaystyle ˉ2^2 = ˉ2\ \timesˉ2 = 4 \ \) because \(\displaystyle ˉ2\) is negative two not minus two!

You are still refusing to recognize the difference between an operator and a sign!

(So I refuse to respond any further to any of this nonsense!)

Go look here.

 

[Dr.Peterson]

Where on earth did I say: "I, lookagain, am rewriting ˉ2² as ˉ1*2²" ???

You are trying to put words in my mouth now?

I have not written any such thing anywhere (because it is utter drivel!)

\(\displaystyle ˉ2^2 = ˉ2\ \timesˉ2 = 4 \ \) because \(\displaystyle ˉ2\) is negative two not minus two!

You are still refusing to recognize the difference between an operator and a sign!

(So I refuse to respond any further to any of this nonsense!)

Go look here.

You are using the raised negative as an indicator of a negative number, and assuming that it is not treated as a separate operator. I don't know how standard that is; can you provide a reliable source to demonstrate that fact?

In any case, since it is not well known, you need to state the meaning of your notation before using it, in order to avoid confusing people.