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General Form of Circle Equation 
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#1
May404, 10:08 AM

P: 8

I'm attempting to write the general form of a circle where I'm given three points, not necessary equally distant from each other: (0,0), (0,8), (6,0)
I need to write the equation of the circle in general form. I've managed to get pretty close to what the answer in the book states, but I'm still off. Since these points are not equally distant from each other, you can't use the midpoint formula (x1+x2)/2, (y1+y2)/2 to get the center of the circle. Is there another method I should be using to find the center so I can determine the radius, and then find the general form for the circle? I'm not looking for the answer, I know what the book says in the answer key, I'm just curious as to what I'm missing, then I can try to take it from there. Thanks! 


#2
May404, 11:54 AM

Sci Advisor
P: 905

the circle equation is
[tex](xa)^2+(yb)^2=r^2[/tex] Plug in each of the three (x,y) coordinate pairs, and you get 3 separate equations. The unknowns are a,b,r. Solve it. 


#3
May504, 10:09 AM

P: 439

[itex]rSin[t]  a; \ rCos[t]  b[/itex] as t goes from 0 to [tex]2\pi[/tex] I may be wrong, so somebody double check it for me. 


#4
May504, 10:46 AM

P: 8

General Form of Circle Equation
Ok, so for the three points you get:
[tex](0a)^2 + (0b)^2 = r^2[/tex] [tex](0a)^2 + (8b)^2 = r^2[/tex] [tex](6a)^2 + (0b)^2 = r^2[/tex] But without knowing a & b, how do you solve for r? 


#5
May504, 12:17 PM

Emeritus
PF Gold
P: 734

You have three equations abd three unknowns. You can, for instance, eliminate r by equation the first two; this gives you one equation with only a and b. You do the same with the first and third equations and get a different equation with a and b. Then you have two new equations on a and b only, which you can solve.
Once you have a and b, you can substitute on one of the original three eqns to get r. 


#6
May604, 01:44 PM

P: 8

Hmmm.. I can see how I can do it for this equation since [tex](0a)^2 = a^2[/tex], so then you just take the square root of both sides to solve for a. But if you have (for example) [tex](1a)^2 = 12a+a^2[/tex], you can't just solve for a can you? You'd be left with [tex]a^22a = 1+[/tex](rest of equation)



#7
May604, 02:13 PM

Emeritus
PF Gold
P: 734

Play with them a little bit... when you eliminate r from first and second, you also eliminate a^2 and b^2, so you only get a constant plus a term in b. Something similar happens combining the first and last eqns.



#8
May604, 02:18 PM

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PF Gold
P: 39,327

[tex](0a)^2 + (0b)^2 = r^2[/tex]
so (1) [tex]a^2+ b^2= r^2[/tex] [tex](0a)^2 + (8b)^2 = r^2[/tex] so (2) [tex]a^2+ b^216b+ 64= r^2[/tex] [tex](6a)^2 + (0b)^2 = r^2[/tex] so (3) [tex]a^2 12a+ 36+ b^2= r^2[/tex] Subtracting 1 from 3, 12a+ 36= 0 or a= 3. Subtracting 1 from 2, 16b+ 64= 0 or b= 4. The a^{2}+ b^{2}= 9+ 16= r^{2} so r= 5. Doesn't come any easier than that! 


#9
May604, 02:58 PM

P: 8

Ah.. I see.. I knew I had to set one equation equal to another and then reduce, but I didn't see the relationship between the three points. I'll crack open the book again when I get off work and try a few more of those problems. Thanks for the guidence!



#10
May1004, 10:42 PM

P: 8

Ok.. I've reworked this, and I'm just about where I was at. Here's what my book shows as the general form of the circle with the following points (same as before):
Points: (0,0), (0,8), (6, 0) Book Answer: [tex]x^2+y^26x8y=0[/tex] When I work this, I get this: [tex](0a)^2+(0b)^2=r^2[/tex] [tex]a^2+b^2=r^2[/tex] [tex](0a)^2+(8b)^2=r^2[/tex] [tex]a^2b^216b+64=r^2[/tex] [tex](6a)^2+(0b)^2=r^2[/tex] [tex]a^2+b^212a+36=r^2[/tex] [tex]a^2+b^2=a^2+b^216b+64[/tex] [tex]16b=64[/tex] [tex]b=4[/tex] [tex]a^2+b^2=a^2+b^212a+36[/tex] [tex]12a=36[/tex] [tex]a=3[/tex] [tex](xa)^2+(yb)^2=r^2[/tex] [tex](03)^2+(04)^2=r^2[/tex] [tex]r=5[/tex] [tex](x3)^2+(y4)^2=5[/tex] [tex]x^23x3x+9+y^24y4y+16=5[/tex] [tex]x^2+y^26x8y+20=0[/tex] As you can see, I'm slightly off, but I don't understand what I've done wrong. Any ideas where I strayed? Thanks again for your help! 


#11
May1104, 12:26 AM

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P: 905




#12
May1104, 04:57 AM

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P: 12,016

Besides, note the swift manipulation done by krab. 


#13
May1104, 09:27 AM

P: 8

Ah.. that's what happens when I work on this late at night. For some reason I figured it didn't need to be squared since I had already determined the radius, but it's clear now.
That is quite a bit simplier manipulation Krab. I've got a second problem I can try this out on, since I don't believe I did it correctly.... I think I need a bigger eraser :) Thanks for working through this with me. I appreciate it! 


#14
May1104, 08:24 PM

P: 47




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