# I was wondering......

joe shmo
Posers and Puzzles 14 Dec '07 02:29
1. joe shmo
Strange Egg
14 Dec '07 02:291 edit
Lets say you have a circle on the cartisian graph

Ill use the numbers I'm toying with now

circle eqtn: (x-2)^2 + (y+1)^2 = 9

I would like to draw a rough sketch of it, but with more than 4 points.

the degree of accuracy depends on the amount of points found

the only way I can think to find more points is to pass a lines through the circle. each line should create two more points of contact.

I passed a line through the point (2,-1) the circles center. i figured the only way to achieve these new points was to randomly change the slope of the line

i used the equation: 5y-2x=1 (picking a random slope)

Now i am currently attempting to solve the system:

5y -2x = 1
(y+1)^2 + (x-2)^2 = 9

except I have been unsuccesful in finding the solutions. I looked through my text, but have found nothing on this particular type of system.

can someone here help me out?
2. 14 Dec '07 03:021 edit
5y - 2x = 1 doesn't pass through the circle. I believe you forgot that the y-coordinate for the center of the circle is -1. If the y-coordinate was 1, then the equations would have two solutions.

If you want four points on that circle, they are (-1, -1), (5, -1), (2, 2), and (2, -4).
3. joe shmo
Strange Egg
14 Dec '07 03:15
Originally posted by twilight2007
5y - 2x = 1 doesn't pass through the circle. I believe you forgot that the y-coordinate for the center of the circle is -1.
ok,....what about the line y = 2/5x - 9/5

this was the line i meant, but i made an error like you said and then i multipled through by 5 to clear fractions.....whoops

i'm mostly interested in how to solve the system
4. joe shmo
Strange Egg
14 Dec '07 03:18
Originally posted by joe shmo
ok,....what about the line y = 2/5x - 9/5

this was the line i meant, but i made an error like you said and then i multipled through by 5 to clear fractions.....whoops

i'm mostly interested in how to solve the system
the ponit of this is to find more than just the four easiest point to find
5. joe shmo
Strange Egg
14 Dec '07 05:20
Originally posted by joe shmo
the ponit of this is to find more than just the four easiest point to find
i went through the book and found something for this system

y = 2/5x - 9/5 the line eqtn

(x-2)^2 + (y + 1)^2 = 9 circle eqtn

I'm hoping I messed the algebra up........here is what i got substituting the linear eqtn into the circle eqtn.

x^2 -4x + 4 + 4/25x^2 - 36/25x + 81/25 + 4/5x -18/5 + 1 = 9

then i multiplied through by 25 to clear fractions and i came up with after simplification

21x^2 -116x -109 = 0

then i used the quadratic formula and came up with a two values

roughly 6.34 and -0.8184 and these are supposed to be the x coordinates?

I know that 6.34 is bogus right of the bat because the x coordinate cant be greater than or equal to 5

so where have i gone wrong?
6. joe shmo
Strange Egg
14 Dec '07 05:55
Originally posted by joe shmo
i went through the book and found something for this system

y = 2/5x - 9/5 the line eqtn

(x-2)^2 + (y + 1)^2 = 9 circle eqtn

I'm hoping I messed the algebra up........here is what i got substituting the linear eqtn into the circle eqtn.

x^2 -4x + 4 + 4/25x^2 - 36/25x + 81/25 + 4/5x -18/5 + 1 = 9

then i multiplied through by 25 to clear ...[text shortened]... t because the x coordinate cant be greater than or equal to 5

so where have i gone wrong?
Sorry for all this.............i found an adition error

in the quadratic standard form ax^2 + bx + c =0

i had the ( a ) coefficient as 21 and it should be 29.....this gives me more plausible answers!!!ðŸ™„
7. TheMaster37
Kupikupopo!
14 Dec '07 07:20
Originally posted by joe shmo
Sorry for all this.............i found an adition error

in the quadratic standard form ax^2 + bx + c =0

i had the ( a ) coefficient as 21 and it should be 29.....this gives me more plausible answers!!!ðŸ™„
It's fun to see someone solve their own problem ðŸ™‚
8. joe shmo
Strange Egg
14 Dec '07 13:55
Originally posted by TheMaster37
It's fun to see someone solve their own problem ðŸ™‚
I'm glad someone is getting a kick over me losing sleep......ðŸ˜µ
9. joe shmo
Strange Egg
14 Dec '07 14:05
I have yet another problem................. take a simple qaudratic......

ex: (x^2 -5x + 4)

It looks to me like the factoring can be solved algebraiclly

like: ab = 4

a + b = -5

except I usually get right back to square one ( another quadratic perhaps with different signs)

im trying to avoid that result............is there any way?ðŸ˜•
10. PBE6
Bananarama
14 Dec '07 14:59
Originally posted by joe shmo
I have yet another problem................. take a simple qaudratic......

ex: (x^2 -5x + 4)

It looks to me like the factoring can be solved algebraiclly

like: ab = 4

a + b = -5

except I usually get right back to square one ( another quadratic perhaps with different signs)

im trying to avoid that result............is there any way?ðŸ˜•
It's fairly straightforward to solve x^2 - 5x + 4 = 0 by inspection (x = -1 or x = -4), but you can also use the quadratic formula:

x = (-b +/- SQRT(b^2 - 4ac)) / (2a)

where a, b and c are the coefficients in the quadratic equation you're trying to solve (in this case, a = 1, b = -5 and c = 4). Just remember that to use the quadratic formula, the quadratic equation must be in standard form.

Try these:

(a) x^2 - 7 + 12 = 0

(b) x^2 + x + 1 = 0

(c) 2x^2 + 3 = 7x
11. joe shmo
Strange Egg
14 Dec '07 15:391 edit
Originally posted by PBE6
It's fairly straightforward to solve x^2 - 5x + 4 = 0 by inspection (x = -1 or x = -4), but you can also use the quadratic formula:

x = (-b +/- SQRT(b^2 - 4ac)) / (2a)

where a, b and c are the coefficients in the quadratic equation you're trying to solve (in this case, a = 1, b = -5 and c = 4). Just remember that to use the quadratic formula, the quadrat ...[text shortened]... andard form.

Try these:

(a) x^2 - 7 + 12 = 0

(b) x^2 + x + 1 = 0

(c) 2x^2 + 3 = 7x
I think your misunderstanding me? I would like to set up a system of equations that will solve for the factors of 4 that will give a b term of -5

like if we let (a) = a factor of 4 and (b) = a factor of 4

just for clarification (x a )(x b)

then the system should be:

ab=4
a + b = -5

I don't want to solve it as a quadratc......... I dont know if what i am asking is possible?

I'm curious of how mathmeticians would have solved this before they new how to solve quadratics...if that makes any sence

I'm trying to see if there is algebra hidden besides the quadratic......and beside guess and check
12. PBE6
Bananarama
14 Dec '07 16:38
Originally posted by joe shmo
I think your misunderstanding me? I would like to set up a system of equations that will solve for the factors of 4 that will give a b term of -5

like if we let (a) = a factor of 4 and (b) = a factor of 4

just for clarification (x a )(x b)

then the system should be:

ab=4
a + b = -5

I don't want to solve it as a quadratc....... ...[text shortened]... rying to see if there is algebra hidden besides the quadratic......and beside guess and check
Ah, what you're talking about there is a Diophantine equation - an equation to be solved using integer values only. These don't have explicit solutions in general, although in special cases it may be possible to derive a formula.

I do recall a method for determining the solutions of a polynomial with rational coefficients, but it involves listing out all the factors of the first and last coefficients, finding their ratios, and then performing long division on the polynomial. I think it's a bit of overkill for a quadratic equation.
13. joe shmo
Strange Egg
14 Dec '07 16:54
Originally posted by PBE6
Ah, what you're talking about there is a Diophantine equation - an equation to be solved using integer values only. These don't have explicit solutions in general, although in special cases it may be possible to derive a formula.

I do recall a method for determining the solutions of a polynomial with rational coefficients, but it involves listing out all th ...[text shortened]... ng long division on the polynomial. I think it's a bit of overkill for a quadratic equation.
ok...... so there is a way

i understand it eisier to simply do it with known methods

Im just trying to conceptualize math a little better......thanks for the tip

thanks
14. wolfgang59
15 Dec '07 12:46
Originally posted by joe shmo
Lets say you have a circle on the cartisian graph

Ill use the numbers I'm toying with now

circle eqtn: (x-2)^2 + (y+1)^2 = 9

I would like to draw a rough sketch of it, but with more than 4 points.

the degree of accuracy depends on the amount of points found

the only way I can think to find more points is to pass a lines through the circle. ...[text shortened]... ext, but have found nothing on this particular type of system.

can someone here help me out?
Keep it simple

You have a circle; all x values lie between -1 and 5. Just plug those x values in and generate as many y values as you want. If you only want integer solutions just plug in x=-1, 0, 1, 2, 3, 4, 5
15. joe shmo
Strange Egg
15 Dec '07 14:291 edit
Originally posted by wolfgang59
Keep it simple

You have a circle; all x values lie between -1 and 5. Just plug those x values in and generate as many y values as you want. If you only want integer solutions just plug in x=-1, 0, 1, 2, 3, 4, 5
i realized that a vertical line passed through the circle would create the same desired effect, with much greater simplicity............after i wasted most of the night doing it this way............ ðŸ™„