Graphs of equations: 3x^2+3y^2-6y-1=0

[AGlas9837]

I'm asked to complete the square to write the equation of the circle in standard form, then use a graphing utility to graph the cirle:

3x^2+3y^2-6y-1=0

I started this problem but couldn't finish. This is what I have so far:

x^2+y^2-2y-1/3=0
(x^2 ?)+(y-2y+1)=1/3+1

I didn't know what to do with just the one term, x^2, on the left! Also, I believe I have to graph 2 equations to actually "graph" the circle. What would those equations be?

 

[o_O]

You don't exactly need 2 equations to graph your circle as the equation you have right there is one already. It's just a matter of putting it into a recognizable form.

As for your work, it is correct. I'm sure you're familiar with the equation of a circle:

\(\displaystyle (x-a)^{2} + (y - b)^{2} = r^{2}\)

Where a and b represent the horizontal and vertical shifts of the centre (0,0) and r representing the radius. If we apply this to your equation, we see that \(\displaystyle a = 0\) (no horizontal shift of the centre) so no worries about that. All you need to worry about now is simplifying y[sup:26aqhnhl]2[/sup:26aqhnhl] - 2y + 1 into the (y - b)[sup:26aqhnhl]2[/sup:26aqhnhl] form.

 

[AGlas9837]

So the equation would be (x+0)^2 + (y-1)^2 = 4/3? You mentioned that I already have an equation but I'm not clear on the graphing part since you can't graph an equation in this form (with y variable). What, exactly, do I put into calculator to graph? Thanks.

 

[stapel]

What, exactly, do I put into calculator to graph?

Back in geometry and also in algebra, you graphed circles in the form (x - h)[sup:3pysez92]2[/sup:3pysez92] + (y - k)[sup:3pysez92]2[/sup:3pysez92] = r[sup:3pysez92]2[/sup:3pysez92], because you'd learned that the center was at the point (h, k) and the radius was the value of r. Just do that. It's much simpler than solving the circle equation for (y - k)[sup:3pysez92]2[/sup:3pysez92] = (whatever), taking the square root of either side (remembering where the "plus-minus" goes), adding k to each side, and then trying to plot points for the two "halves" of y. :shock:

Just do what you learned in your earlier classes! :wink:

Eliz.

 

[AGlas9837]

Elizabeth,

You're giving me way too much credit, I'm afraid because I DON'T remember a lot of my geometry and algebra and I'm still pulling my hair out here trying to graph this stupid equation. I STILL don't understand what equation to plug into my calculator to graph this. Sorry!

 

[o_O]

You have this:

(x+0)^2 + (y-1)^2 = 4/3

Recall the formula of the circle:

\(\displaystyle (x - a)^{2} + (y - b)^{2} = r^{2}\)

(a, b) gives you the centre of the circle and r is the radius. So ...

You know where the centre is (0, 1) and has a radius of \(\displaystyle \sqrt{4/3}\).

If you really need to graph it on your calculator, just solve for y as stapel explained.

 

[PAULK]

I'm afraid because I DON'T remember a lot of my geometry and algebra and I'm still pulling my hair out here trying to graph this stupid equation. I STILL don't understand what equation to plug into my calculator to graph this.

You're OK, as far as this goes. Now proceed:

x^2 + (y - 1)^2 = 4/3

Solve for y:

(y - 1)^2 = 4/3 - x^2

y - 1 = +- sqrt(4/3 - x^2) << I can't make 'plus-or-minus' characters.

y = 1 +- sqrt(4/3 - x^2)

ok. Now use your graphing calculator to plot BOTH OF THESE:
(A good one will plot two graphs for you if you smile sweetly.)

y = 1 + sqrt(4/3 - x^2)
y = 1 - sqrt(4/3 - x^2)

and you put them together with your own pencil.

 

[AGlas9837]

Perfect! God Bless!!