math question
03-05-2016, 03:52 PM
Post: #1
 Don Shepherd Senior Member Posts: 745 Joined: Dec 2013
math question
or maths, if you are not in the US

I am trying to graph circles on my TI-84 plus graphing calculator. I know that the formula for a circle is (x-h)^2 + (y-k)^2 = r^2 where the center of the circle is (h,k) and the radius is r, but to actually graph the circle I need two formulas (one for the upper part, one for the lower) in the form y1= and y2=, using variables x, h, k, and r.

I have found several formulas on the Internet, but none give me the results that I would expect.
03-05-2016, 04:11 PM
Post: #2
 SlideRule Senior Member Posts: 1,324 Joined: Dec 2013
RE: math question
How do I graph a circle on a TI graphing calculator?
Functions can be graphed on TI graphing calculators in the form Y in terms of X.
In order to graph a circle, the function for a circle would need to be solved for Y, ie to graph the unit circle (x2+y2=1), first, solve for y, and then input the results into the Y= Editor

source: "Graphing a Circle on a TI Graphing Calculator. - Knowledge Base by Texas Instruments - US and Canada"

BEST!
SlideRule
03-05-2016, 04:57 PM
Post: #3
 Manolo Sobrino Member Posts: 179 Joined: Dec 2013
RE: math question
Why don't you try a parametric plot?

Mode->Par

X_{1T} = h + r*cos(T)

Y_{1T} = k + r*sin(T)

Zoom->ZSquare or ZDecimal or ZInteger

If you insist on using Cartesian coordinates, then

Mode->Func

Y_{1} = k +\sqrt{r^2-(x-h)^2}

Y_{2} = k -\sqrt{r^2-(x-h)^2}

Zoom->ZSquare or ZDecimal or ZInteger
03-05-2016, 05:46 PM
Post: #4
 Don Shepherd Senior Member Posts: 745 Joined: Dec 2013
RE: math question
Thanks SlideRule and Manolo, that's exactly what I was looking for.
03-07-2016, 08:38 AM
Post: #5
 klesl Member Posts: 78 Joined: Mar 2016
RE: math question
You can use following application for this task
https://education.ti.com/en/us/software/...icgraphing

There are 4 shapes - circle, ellipse, hyperbola and parabola
03-07-2016, 09:08 PM
Post: #6
 Don Shepherd Senior Member Posts: 745 Joined: Dec 2013
RE: math question
Thanks Klesl, that is a good reference.
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