The fun and joy of a summer calc class has brought me to this question that wants me to graph a circle equation on a calculator, which I've never really done or don't remember. The equation (in general form) is 4x^2 +4y^2 - 4x +24y - 63 = 0. Any idea how to get this set equal to y? I keep trying, but I can't get it by itself because of the pesky squares. Any help at all would be appreciated.

Its impossible to graph the full circle with one equation of the form y=f(x) because a circle has 2 y-values for each x-value, but you can do it with two equations, one for the top half of the circle as y=f(x) and then the bottom half as y=-f(x). To find the explicit form of f(x) in this case you proceed as follows.

We have 4x^2 +4y^2 - 4x +24y - 63 = 0 so 4y^2+24y=-4x^2+4x+63 then factoring gives 4(y^2+6y)=4(63/4+x-x^2). Dividing by 4 you get y^2+6y=(63/4+x-x^2). One the left side of the equation you can complete the square by adding 9, remembering to add 9 to the right side also so you have y^2+6y+9=(63/4+x-x^2)+9 then factoring the left side and some computation on the right side gives (y+3)^2=(99/4+x-x^2). Then taking the square root of both sides gives y+3=(99/4+x-x^2)^1/2 so the final equation is y=((99/4+x-x^2)^1/2)-3. This is just the top half of the circle as explained above, with the bottom half being y=-[((99/4+x-x^2)^1/2)-3], just the negative of the top half equation.

well, circles arent functions, so im not sure if your calculator can graph it

Thanks a lot danreil, and yeah, my calculator can graph nonfunctions, as long as it's expressed in terms of y. y^2=x can be expressed as long as you square both sides.

but if you square root to make y the subject, youll get
y=x^(1/2)

x cannot be negative, so youll only have half the function.

or so i think

You need to use the conics program on your calculator or split it into two functions. I wouldn't recommend you do that though. Does your calculator have a conics program?

d00d just make a parametric equation out of it and graph that, TI-83's can do that for a fact

4x^2 +4y^2 - 4x +24y - 63 = 0

4x^2-63=-(4y^2+24Y)

x^2+15.75=y^2+6y

(x^2+15.75)^(1/2)=(y^2+6y)^(1/2)

x+3.969=2.45y

y=(x+3.70)/2.45

http://i255.photobucket.com/albums/hh150/XtomJames/Calculation2.jpg

4x^2 +4y^2 - 4x +24y - 63 = 0

4x^2-63=-(4y^2)+24Y

x^2-15.75=y^2+6y

(x^2-15.75)^(1/2)=(y^2+6y)^(1/2)

x-3.969=-2.45y

y=(x-3.70)/-2.45

http://i255.photobucket.com/albums/hh150/XtomJames/Calculation.jpg

Wow, old thread, but don't listen to this because its incorrect. Your picture isn't even a circle and almost every step you wrote has a mistake in it.

Uh no, a circle equation will not graph as a circle it will always graph as a hyperbalic structure.

The steps taken are balancing the equation and are correct.

What the hell are you talking about, of course an equation for a circle graphs as a circle, they're equivalent statements.

The original equation is 4x^2 +4y^2 - 4x +24y - 63 = 0 but then you write 4x^2-63=-(4y^2)+24Y. You forgot about the 4x term in going from the original equation to 4x^2-63=-(4y^2)+24Y, which is a mistake.

Also, ignoring that for the moment, you also have written (x^2+15.75)^(1/2)=(y^2+6y)^(1/2) then x+3.969=2.45y, which is blatantly incorrect if you even have a basic knowledge of algebra.

Ok yes I guess you are right I completely looked over the 4x, by accident. However, even then in a standard xy graph it won't graph a circle because to graph a circle you need an F(x,y) graph instead of an F(x) graph in the graphing calculator.

4x^2 +4y^2 - 4x +24y - 63 = 0

4x^2 +4y^2+24y - 63 = 4x

4x^2 +4y^2+24y= 4x+63

4y^2+24y= 4x+63-4x^2

(4y^2+24y= 4x+63-4x^2 )/4

(y*y)+6y= x+15.75-(x*x)

y^2=(x+15.75-(x^2))-6y

(y^2=(x+15.75-(x^2))-6y)/y

y=-x^2+x+9.75

(y^2=(x+15.75-(x^2))-6y)/y

y=-x^2+x+9.75

In this step, the last step you have, you did the algebra incorrectly. After dividing by y in the top equation you should have y=(x/y)+(15.75/y)-((x^2)/y)-6. You made your mistake because in the top equation, when you divided by y, you divided term-by-term like you should, but you only divided the y terms by y, for instance making 6y into 6, but you neglected to divide the x terms by y, for instance you have x instead x/y.

Er, no, when dividing a whole string, the components are removed. You aren't dividing each individual seperately.

Lets see if I can explain this better. Say you have the equation y^2=x^2-6y. Now say y=2, then we get that x=4. Now divide by y, just like we did earlier. According to your method you get y=x^2-6 and according to my method you get y=((x^2)/y)-6.

Since in the original equation, if you put in y=2 and x=4 you get that both sides are equal, then if your method is correct, if we put in y=2 and x=4 in your equation,y=x^2-6, then both sided should still be equal. Then since y=2 and x^2-6=10 when x=4, we see that both sides are not in fact equal, so there was a mistake made.

Now take my equation, y=((x^2)/y)-6. y=2 on one side and ((x^2)/y)-6=2 when x=4 and y=2. Thus both sides of the equation are equal and this method is the correct one in this case. You can try this with any other equation you want by substituting numbers for the variables and see this method will always work. Its also the way you should learn to divide expressions in algebra 1.

http://img176.imageshack.us/img176/340/dist2lg3.jpg

10 generations of pox!

http://img176.imageshack.us/img176/340/dist2lg3.jpg

Is this clear enough?