math help

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madah123
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math help

Post by madah123 »

i have amath question i couldnt solve in 3 hours.
its to find the equation of acircle that touch the three points
(2,0) -(0،-12) - (6،8)
by using the general form of the equation
x^2+Y^2+ax+by+c=0

each time i tried to solve it .
the equation applied to the first and the second point but not the third

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madah123
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Re: math help

Post by madah123 »

I tried solving it like this:
-.- the point (2،0) is on the circle .-. it fulfils the equation
x^2+y^2+ax+by+c=0 .-.
2^2+0^2+2a+0y+c=0
4+2a+c=0
2a+c=-4 ---->1
--------------------------------------------------------------------------------------------------------------
-.- the point (0،-12) is on the circle .-. it fulfils the equation
x^2+y^2+ax+by+c=0 .-.
0^2+-12^2+0+-12b+c=c
144-12b+c=0
-12b+c=-144 * -1
12b-c=144 ------->2
---------------------------------------------------------------------------------------------------------------------
-.- the point (6،8) is on the circle .-. it fulfils the equation
x^2+y^2+ax+by+c=0 .-.

6^2+8^2+6a+8b+c=0
100+6a+8b+c=0
6a+8b+c=-100 --------->3
------------------------------------------------------------------------------------------------------------------------
then i tried to combine the equations together still i couldnt get the right answer

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NuclearSilo
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Re: math help

Post by NuclearSilo »

it's easy, it's just a system of 3 equation:
answer: a=-71 b=23.5 c=138
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madah123
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Re: math help

Post by madah123 »

Oh my answers were kinda close i got
a= -74.375
b=24.0625
c=144.75
Last edited by madah123 on Tue Mar 10, 2009 12:31 pm, edited 1 time in total.

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takolin
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Re: math help

Post by takolin »

x²+y²+ax+by+c=0

Points: (2,0) (0,-12) and (6,8)

Fill in each point in the equation.

2²+0²+2a+0b+c = 0
0²+(-12)²+0a-12b+c=0
6²+8²+6a+8b+c=0

"Reduce" them.

2a+c+4=0
-12b+c+144=0
6a+8b+c+100=0

Now you need to solve those 3 equations in order to know what a,b and c is. And you'll have the equation for your circle.

The most easy solution for the 3 equations would be a making a matrix and reduce it.
But I can't be arsed doing it without my calculator.

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madah123
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Re: math help

Post by madah123 »

takolin wrote:x²+y²+ax+by+c=0

Points: (2,0) (0,-12) and (6,8)

Fill in each point in the equation.

2²+0²+2a+0b+c = 0
0²+(-12)²+0a-12b+c=0
6²+8²+6a+8b+c=0

"Reduce" them.

2a+c+4=0
-12b+c+144=0
6a+8b+c+100=0


Now you need to solve those 3 equations in order to know what a,b and c is. And you'll have the equation for your circle.

The most easy solution for the 3 equations would be a making a matrix and reduce it.
But I can't be arsed doing it without my calculator.

And how is that diffrent than what i did ?
I have no problems with making the 3 equation i have problems with combining them

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takolin
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Re: math help

Post by takolin »

What do you mean with combine.

You found the 3 equations like I did.
You just failed to solve them. I can't help you with that one, ask your math-teacher.

http://www.easycalculation.com/matrix/m ... lgebra.php

A nice shortcut if you don't have a calculator near you. It won't teach you the basics of solving equations though, but it makes things quicker if you know how to do so.



Edit:

A small help in how to solve it.

2a+c=-4
<=>
a = (-4-c)/2

12b -c = 144
<=>
b = (144+c)/12

Now substitute a and b in the 3rd equation.
Now you only have 1 parameter (c).
Find a solution for c.
Find a and b by filling in c in the 2 equations I gave you.

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madah123
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Re: math help

Post by madah123 »

I will post exactly what i did so tell where i did wrong
2a+c=-4
+
12b-c+144
=
2a+12b=140 *3
=
6a+36b=420
-
6a+8b+c=-100
=28b-c=520
+
2a+c=-4
=2a+28b=516 *3
=6a+84b=1584
--------------------------------------
6a+8b+c=-100
+
12b-c=144
=
6a+20b=44
---------------------------------------------
=6a+84b=1584
-
+a+20b=24
=
64b=1504
b=24.0625

12b-c=144
288.75-c=144
c=144.75
2a+c=-4
2a+144.75=-4
a=74.375
_____________________________________________________________________
i rechecked it alot of times i cant see where i went wrong

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NuclearSilo
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Re: math help

Post by NuclearSilo »

woa woa, what's with this wall of code.
Try simply add/sub the equation 3 and 1, 3 and 2 you'll get 2 equation with a and b. After that is a piece of cake.
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takolin
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Re: math help

Post by takolin »

NuclearSilo wrote:woa woa, what's with this wall of code.
Try simply add/sub the equation 3 and 1, 3 and 2 you'll get 2 equation with a and b. After that is a piece of cake.


My method is easier imo.
In the end you have 1 equation with 1 unknown figure.

Or just use a matrix. My calculator solves those as well.

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Re: math help

Post by Riptide »

Seriously the key to math problems is to think simple, your thinking to complex. Keep it simple and focus on the problem before you. Don't start thinking to deep.
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DonBebek
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Re: math help

Post by DonBebek »

Too bad I absolutely despise math-.-

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JustaDude
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Re: math help

Post by JustaDude »

You need to study more dude, that was one easy question indeed.
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Re: math help

Post by bryanvan »

Riptide wrote:Seriously the key to math problems is to think simple, your thinking to complex. Keep it simple and focus on the problem before you. Don't start thinking to deep.


My teachers tells me everything that I have that prob XD that I think way to difficult, thats why mostly I got the most difficult questions right and and the really easy ones wrong.
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Riptide
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Re: math help

Post by Riptide »

bryanvan wrote:
Riptide wrote:Seriously the key to math problems is to think simple, your thinking to complex. Keep it simple and focus on the problem before you. Don't start thinking to deep.


My teachers tells me everything that I have that prob XD that I think way to difficult, thats why mostly I got the most difficult questions right and and the really easy ones wrong.


Though problems require easy solutions!
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