At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
Are you familiar with what sort of shape you'll get from an equation of this form?
Not this one
Okay. Equations of the form \((x - h)^2 + (y - k )^2 = r^2\) are considered circles. The center of the circle is (h, k), and the radius is r. So, you can usually make comparisons between the 'standard form'and the given equation to figure out your center + radius, and graph from that. :)
will the center b (0,-3)?
Yes, that is correct.
So then this will be a circle intersecting at ±3 all round
Hmm... well, the circle simply has a radius of three, so basically all the points that are a distance of 3 from (0, -3) are points on your graph. It makes sense if you consider r the distance between the center and all the points on the circle that are 3 units away. :)
The exact center of the circle will be at (0, -3), like this: |dw:1352107269477:dw|
Since the radius is 9 and the √9=3
Our radius is determined as r^2 = 9 => r = 3. :) So, we can draw in all the points around (0, -3) that are a distance of 3 units away for the circle.
|dw:1352107502707:dw| So, we can pick out a few simple ones that line up on the axes, and then we have to estimate from there. My graph isn't drawn particularly well, so it doesn't look very nice, but you should be able to get a circle. :P
So the points are (0,0) (3,-3) (0,-6) (-3,-3)?
Yes, those are a few points on the graph of the circle. If you start from your center, it is easiest to move horizontally or vertically by the same distance as the radius to find four points to draw the rest of the graph from.
The process of graphing the circle basically goes like this: 1) Find your center-point and radius 2) Plot the center-point on your graph 3) Plot those four points at a distance of one radius away and draw in the circle from there as best you can. Usually teachers will be forgiving as long as you have those four points marked and your circle looks somewhere close. :P
You've been so much helpful
I have about 6 more qtns to go. I'm staying awake till class time which is 7am
I'm glad to be able to help! :) Although, I will have to get off. I should really be sleeping...
Sure. Thank a million.
Good night and sound sleep.