How to proceed with this?

allegansveritatem said:
right, simplified is beautiful, to paraphrase somebody.
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@allegansveritatem, If you go back to reply #3 and use the links then you will see:
If \(|b|<2\sqrt 2\) then the line will intersect the circle in two points. Try changing \(b=2\).
If \(|b|=2\sqrt 2\) then the line will intersect the circle exact one point. Look at both links.
If \(|b|>2\sqrt 2\) then the line will intersect the circle in no point. Try changing \(b=4\).
 
allegansveritatem said:
right, simplified is beautiful, to paraphrase somebody. I will have to look at it again. I worked it out again today and got strung out on something to do with the variable (b) and had a photo I was going to upload and a question but then I realized my mistake and so nothing needs be said on the matter. I still haven't clearly understood on a procedural level how to do these systems problems where there is more than one solution per variable. I need to go back and study the chapter again.
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Yes. I said one error was careless. The other is not. The other requires attention to definitions.

[MATH]\sqrt{x^2} = |x|.[/MATH]
[MATH]\therefore 9 > x^2 \implies \text {exactly what about } x.[/MATH]
By the way, I have been approaching this strictly algebraically. Others have been approaching this through analytic geometry. Both approaches work. Use what works for you.
 
JeffM said:
Yes. I said one error was careless. The other is not. The other requires attention to definitions.

[MATH]\sqrt{x^2} = |x|.[/MATH]
[MATH]\therefore 9 > x^2 \implies \text {exactly what about } x.[/MATH]
By the way, I have been approaching this strictly algebraically. Others have been approaching this through analytic geometry. Both approaches work. Use what works for you.
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that x is greater than three?
 
Jomo said:
Here is how I would do this problem.
The circle is centered at the origin. The line y = x + 0 will cross the circle twice.
If we let b differ from 0 by just a little, then the line will cross the circle twice.
If we let b differ from 0 by a bit more then the line will cross the circle twice.
At some point if b differs from 0 enough then the line will cross the circle in exactly one point.

Now we need to remove words like enough, a little and a bit more!

The slope of all lines of the form y=x+b is 1.

All we need to do is find out the two points where the slope of the tangent line to the circle is 1. Where does these two tangent lines cross the y-axis. Those are your two b* values. If b (in y=x+b) is in-between the two b* values you just found the line will cross the circle twice. If b equals one of the two b*values you just found then the line will cross the circle in one place.
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so, I tried this today and came up with this:
aug252.PNG
Here is the calculator scrren I was working with:
aug251.PNG

the line running under and parallel to the x axis is y=-sqrt 2. I entered that to establish that x-2.9 (approx) intersects the circle at (sqrt2,-sqrt2).
 
JeffM said:
[MATH]9 > x^2 \implies 3 > |x| \implies -3 < x < 3.[/MATH]
[MATH]x = -4 \implies x^2 = 16 \not < 9.[/MATH]
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JeffM said:
[MATH]9 > x^2 \implies 3 > |x| \implies -3 < x < 3.[/MATH]
[MATH]x = -4 \implies x^2 = 16 \not < 9.[/MATH]
Click to expand...
I am not sure what you are saying here. I had three objectives given me by the author of this conundrum which were: a) what values of b would produce one solution, b)two solutions and c) no solutions. The discriminant is sqrt-4B^2+32. If we say that x-4=y then we can get no real solutions and thus 4 would serve our turn. No? I'm not sure if I posted the full problem in my original post--the text ran from one page in the book to the next. a)and b) were asked for on one page and c) on the other.
 
allegansveritatem said:
I am not sure what you are saying here. I had three objectives given me by the author of this conundrum which were: a) what values of b would produce one solution, b)two solutions and c) no solutions. The discriminant is sqrt-4B^2+32. If we say that x-4=y then we can get no real solutions and thus 4 would serve our turn. No? I'm not sure if I posted the full problem in my original post--the text ran from one page in the book to the next. a)and b) were asked for on one page and c) on the other.
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You were correct when, back in post 14, you got the discriminant

[MATH]-4b^2 + 32 = 4(8 - b^2).[/MATH]
One real root will occur if the discriminant is zero. Here is where you made your first (probably careless) error.

[MATH]4(8 - b^2) = 0 \implies 8 - b^2 = 0 \implies \\ b^2 = 8 \implies b = \pm \sqrt{8} = \pm \sqrt{4 * 2} = \pm 2\sqrt{2}.[/MATH]You gave as your answer positive 8: you forgot to take the square root and so forgot that there was both a positive and negative answer.

Two real roots will occur if the discriminant exceeds zero. Here you made your more subtle error.

You gave as your answer [MATH](- \infty,\ 8)[/MATH].

But it is not true that

[MATH]8 - (-3)^2 = 8 - (+9) = 8 - 9 = - 1 > 0[/MATH].

In general

[MATH]x^2 < y > 0 \implies - \sqrt{y} < x < \sqrt{y}.[/MATH]
 
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allegansveritatem said:
I am not sure what you are saying here. I had three objectives given me by the author of this conundrum which were: a) what values of b would produce one solution, b)two solutions and c) no solutions. The discriminant is sqrt-4B^2+32. If we say that x-4=y then we can get no real solutions and thus 4 would serve our turn. No? I'm not sure if I posted the full problem in my original post--the text ran from one page in the book to the next. a)and b) were asked for on one page and c) on the other.
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It may help if you back off and look from a different perspective.

Here is a graph of the discriminant, [MATH]-4b^2+32[/MATH] (taking b as the horizontal axis):

1598450103807.png

For what values of b is this positive? negative?
 
Beer soaked query follows.
allegansveritatem said:
I am not sure what you are saying here. I had three objectives given me by the author of this conundrum which were: a) what values of b would produce one solution, b)two solutions and c) no solutions. The discriminant is sqrt-4B^2+32. If we say that x-4=y then we can get no real solutions and thus 4 would serve our turn. No? I'm not sure if I posted the full problem in my original post--the text ran from one page in the book to the next. a)and b) were asked for on one page and c) on the other.
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From which book did you get your conundrum?
 
JeffM said:
You were correct when, back in post 14, you got the discriminant

[MATH]-4b^2 + 32 = 4(8 - b^2).[/MATH]
One real root will occur if the discriminant is zero. Here is where you made your first (probably careless) error.

[MATH]4(8 - b^2) = 0 \implies 8 - b^2 = 0 \implies \\ b^2 = 8 \implies b = \pm \sqrt{8} = \pm \sqrt{4 * 2} = \pm 2\sqrt{2}.[/MATH]You gave as your answer positive 8: you forgot to take the square root and so forgot that there was both a positive and negative answer.

Two real roots will occur if the discriminant exceeds zero. Here you made your more subtle error.

You gave as your answer [MATH](- \infty,\ 8)[/MATH].

But it is not true that

[MATH]8 - (-3)^2 = 8 - (+9) = 8 - 9 = - 1 > 0[/MATH].

In general

[MATH]x^2 < y > 0 \implies - \sqrt{y} < x < \sqrt{y}.[/MATH]
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I will have to go back over what I did and relate it to this post before I can reply, which I will do and get back. I have to be in a certain groove before I can function at all in this subject.At six in the morning I can just barely tie my shoes. But I like to read these posts and think about them as I go about later in the day. Sometimes a flash of what I might manically call insight occurs.
 
JeffM said:
You were correct when, back in post 14, you got the discriminant

[MATH]-4b^2 + 32 = 4(8 - b^2).[/MATH]
One real root will occur if the discriminant is zero. Here is where you made your first (probably careless) error.

[MATH]4(8 - b^2) = 0 \implies 8 - b^2 = 0 \implies \\ b^2 = 8 \implies b = \pm \sqrt{8} = \pm \sqrt{4 * 2} = \pm 2\sqrt{2}.[/MATH]You gave as your answer positive 8: you forgot to take the square root and so forgot that there was both a positive and negative answer.

Two real roots will occur if the discriminant exceeds zero. Here you made your more subtle error.

You gave as your answer [MATH](- \infty,\ 8)[/MATH].

But it is not true that

[MATH]8 - (-3)^2 = 8 - (+9) = 8 - 9 = - 1 > 0[/MATH].

In general

[MATH]x^2 < y > 0 \implies - \sqrt{y} < x < \sqrt{y}.[/MATH]
Click to expand...
I went back to the beginning and redid the algebra and got this
discrim1.PNG:
Feels good to me.
 
To OP, I have absolutely no idea why you would choose to solve for y in x^2 + y^2 = 4 when it would involve sorts and +/- signs when you could have just substituted x+b for y as y=x+b was given!

I had enough of this post. Below are three ways to solve this problem.

1) x^2 + (x+b)^2 = 4
2x^2 + 2bx + b^2 = 4
2x^2 + 2bx + (b^2 - 4) = 0
x= [-2b +/- sqrt(4b^2 - 4*2*(b^2-4)]/4 = [-b +/- sqrt(b^2 - 2(b^2-4)]

Let's consider the discriminant. b^2 - *2*(b^2-4) = b^2 -2b^2+8 = -b^2 + 8

When is -b^2+8=0? When b =/- 2sqrt(2)
That is when you have one point of intersection. Think about how you will get two points of intersection and no points of intersection
Go and find the point which is easy to do.

2)x^2 + y^2 = 4 => y = +/-sqrt(4-x^2)
So +/-sqrt(4-x^2) = x+b
Then 4-x^2 = (x+b)^2 = x^2 + 2bx + b^2
Then 2x^2 + 2bx + (b^2-4) = 0. Now continue as above.

3)To have one point of the intersection the line y=x+b must be tangent to the x^2 + y^2 =4. The slope of y=x+b is 1 and the derivative of a circle is known to be y'= -x/y. Hence -x/y=1 or y=-x
Then x^2 + y^2 = 4 becomes x^2 + (-x)^2 = 4. Then x^2 = 2. So x=+/- sqrt(2)
This gives us the points (sqrt(2), - sqrt(2)) and (-sqrt(2), sqrt(2))

1st point: y = x + b so -sqrt(2) = sqrt(2) + b. So b = -2sqrt(2)
2nd point: sqrt(2) = -sqrt(2) + b. So b = sqrt(2).

if -sqrt(2) < b < sqrt(2) you have 2 points of intersection. If b = +/- sqrt(2) you have one point of intersection. Otherwise you have no points of intersection.


Comments: In my opinion the 3rd way is the way to do this problem. It is simple and shows a good understanding of what is going on.
If you have not YET reached the level of maturity to do this problem by method 3 you should then use method 1. I would never consider using method 2 HOWev
 
Jomo said:
To OP, I have absolutely no idea why you would choose to solve for y in x^2 + y^2 = 4 when it would involve sorts and +/- signs when you could have just substituted x+b for y as y=x+b was given!

I had enough of this post. Below are three ways to solve this problem.

1) x^2 + (x+b)^2 = 4
2x^2 + 2bx + b^2 = 4
2x^2 + 2bx + (b^2 - 4) = 0
x= [-2b +/- sqrt(4b^2 - 4*2*(b^2-4)]/4 = [-b +/- sqrt(b^2 - 2(b^2-4)]

Let's consider the discriminant. b^2 - *2*(b^2-4) = b^2 -2b^2+8 = -b^2 + 8

When is -b^2+8=0? When b =/- 2sqrt(2)
That is when you have one point of intersection. Think about how you will get two points of intersection and no points of intersection
Go and find the point which is easy to do.

2)x^2 + y^2 = 4 => y = +/-sqrt(4-x^2)
So +/-sqrt(4-x^2) = x+b
Then 4-x^2 = (x+b)^2 = x^2 + 2bx + b^2
Then 2x^2 + 2bx + (b^2-4) = 0. Now continue as above.

3)To have one point of the intersection the line y=x+b must be tangent to the x^2 + y^2 =4. The slope of y=x+b is 1 and the derivative of a circle is known to be y'= -x/y. Hence -x/y=1 or y=-x
Then x^2 + y^2 = 4 becomes x^2 + (-x)^2 = 4. Then x^2 = 2. So x=+/- sqrt(2)
This gives us the points (sqrt(2), - sqrt(2)) and (-sqrt(2), sqrt(2))

1st point: y = x + b so -sqrt(2) = sqrt(2) + b. So b = -2sqrt(2)
2nd point: sqrt(2) = -sqrt(2) + b. So b = sqrt(2).

if -sqrt(2) < b < sqrt(2) you have 2 points of intersection. If b = +/- sqrt(2) you have one point of intersection. Otherwise you have no points of intersection.


Comments: In my opinion the 3rd way is the way to do this problem. It is simple and shows a good understanding of what is going on.
If you have not YET reached the level of maturity to do this problem by method 3 you should then use method 1. I would never consider using method 2 HOWev
Click to expand...
A while ago you suggested I go to reply 13 and try that. I did just that and posted the results. Maybe you did not see that post so I post it again here:
aaa1.PNG
aaa2.PNG

Is this something like what you had in mind?
 
allegansveritatem said:
A while ago you suggested I go to reply 13 and try that. I did just that and posted the results. Maybe you did not see that post so I post it again here:
View attachment 21278
View attachment 21279

Is this something like what you had in mind?
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Yes yes, I saw your post. It was just that you went back to the other method.
I hope that you understand all three methods.
If you have any question then please ask.
 
Jomo said:
Yes yes, I saw your post. It was just that you went back to the other method.
I hope that you understand all three methods.
If you have any question then please ask.
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I think I am getting a wholer (?) picture of the situation. Thanks for your suggestions.
 
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