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 May 5th, 2007, 02:24 PM #1 Newbie   Joined: May 2007 Posts: 1 Thanks: 0 graphing help how do I sketch the graph of (x-4)2+(y+3)2=25 the italic 2's mean squared I'm completely retarded and don't know how to do superscript 2's.
 May 5th, 2007, 04:14 PM #2 Senior Member   Joined: Dec 2006 Posts: 1,111 Thanks: 0 So you're trying to graph: (x-4)^2+(y+3)^2=25 That's the graph of a circle of radius 5 centered at the point 4,-3 If you're trying to graph it in rectangular coordinates, forget it. Go to polar coordinates. In polar coordinates, this is equal to the function r=5 .
 May 6th, 2007, 03:03 PM #3 Senior Member   Joined: Nov 2006 From: I'm a figment of my own imagination :? Posts: 848 Thanks: 0 You can graph it in rectangular form, but only as two separate equations, which are somewhat messy. Following are the steps converting it into two functions of x. (x-4)²+(y+3)²=25 (y+3)²=25-(x-4)² y+3=±√[25-(x-4)²] y=-3±√[25-(x-4)²] You would input the + and - as two separate functions, i.e., y_1=-3+√(25-(x-4)²) y_2=-3-√(25-(x-4)²) In polar coordinates, it is not quite r=5, because its center is not the origin. Unfortunately, I do not know how to perform translations in polar co-ordinates. A better approach might be a parametric, rather than polar, graph. We now want to express x and y in terms of a dummy variable t. To simplify this, we define x'=x-4 and y'=y+3. We now have x'=5cos(t) and y'=5sin(t) as our equation. Now, because x=x'+4 and y=y'-3 (by simple rearrangement), x=5cos(t)+4 and y=5sin(t)-3, a much more pleasing system of equations than that required by writing y as a function of x. I have used several symbols here that cannot simply be typed (including the superscript 2's that you were commenting about earlier). To find out how I did this, see the sticky at the top of this forum.[/url]
May 6th, 2007, 06:05 PM   #4
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Quote:
 In polar coordinates, it is not quite r=5, because its center is not the origin. Unfortunately, I do not know how to perform translations in polar co-ordinates.
True, I hadn't thought of that. However, you can just convert rectangular to polar by using the fact that x = r*cos(θ) and y = r*sin(θ) .

Hence, after multiplying everything out, canceling, and applying trig identities, the equation in polar coordinates is:

r = 8cos(θ) - 6sin(θ)

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