Circle Geometry

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Probability

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I have an equation; 3y = - 5x + 14. I want to find the coordinates of any points.

This is what I think I have understood;

(x + 3)^2 + (y - 4)^2 = 17

(x + 3)^2 + ((-5x)/(3) + (14)\(3) - 4))^2 = 17

x^2 + 6x + 9 + ((-5x)/(3) + (14)\(3) - 4))^2 = 17

The problem is that I don't know what to do with the math in brackets?

Do I have to multiply it all out and if so what method?

Do I have to change the denominators so that - 4 is over a denominator?

Do I have to multiply the left hand side by 3 to get rid of the denominator 3?

Any help thanks
 
Probability said:
I have an equation; 3y = - 5x + 14. I want to find the coordinates of any points.

This is what I think I have understood;

(x + 3)^2 + (y - 4)^2 = 17

(x + 3)^2 + ((-5x)/(3) + (14)\(3) - 4))^2 = 17

x^2 + 6x + 9 + ((-5x)/(3) + (14)\(3) - 4))^2 = 17

The problem is that I don't know what to do with the math in brackets?

Do I have to multiply it all out and if so what method?

Do I have to change the denominators so that - 4 is over a denominator?

Do I have to multiply the left hand side by 3 to get rid of the denominator 3?

Any help thanks
Click to expand...

Please post the actual problem - verbatim - for us to understand.

As posted, the problem does not make sense!!

You start with an equation of a straight-line ( linear equation) - then suddenly jump into an equation for a circle.

Probably you are trying to find points of intersection. But before we spend more time "guessing" - please post the original problem.
 
Probability said:
I have an equation

3y = - 5x + 14

I want to find the coordinates of any points.
Click to expand...

This is a linear equation

y = -(5/3)x + 14/3

Substitute any Real number for x and calculate y; you may obtain as many points on the line as you desire.



x^2 + 6x + 9 + ((-5x)/(3) + (14)\(3) - 4))^2 = 17

The problem is that I don't know what to do with the math in brackets?
Click to expand...

You have more closing parentheses than open parentheses.

You seem to have posted only part of the exercise. Please post the entire exercise.

Cheers :cool:
 
Ok I'll start over.

I was given an equation and asked to find the centre and radius. The equation represents a circle. x^2 + y^2 +6x - 8y + 8 = 0

By completing the square I got;

Completing the square

x2 + 6x + y2 – 8y + 8 = 0
x2 + 2px = (x + p)2 – p2
x2 + 6x = (x + 3)2 – 32 = (x + 3)2 + 9
y2 – 8y = (y – 4)2 – 42 = (y – 4)2 + 16
(x + 3)2 + (y – 4)2 + 9 + 16 – 8 = 17

The circle centre ( - 3, 4) and the radius r2 = 17
r = √17

I am then asked to find the coordinates of any points at which this circle intersects the line 3y = - 5x + 14.

So I ended up with;

x^2 + 6x + 9 + (-5/3x + 10/3) = 17

Factors, (3, 3) and (-5/3, - 7/3)

If the above turns out correct then I have established that the equation has two real roots.

If I am incorrect I would appreciate any help.
 
mmm4444bot said:
I did not check the square completion, but your result above would be correct if you were to remove the part in red.




These results are correct.






14/3 - 4 does not equal 10/3.
Click to expand...

There must be a mathematical way then to make all the denominators the same if that is the right way forward from this point I am not sure?

so, - 5x/3 + 14/3 - 4/1
 
Probability said:
There must be a mathematical way then to make all the denominators the same if that is the right way forward from this point I am not sure?

-5x/3 + 14/3 - 4/1
Click to expand...


Yes -- there is a mathematical way; it's called "getting a common denominator".



We convert the fraction 4/1 to an equivalent number of thirds by multiplying 4/1 by 3/3.

\(\displaystyle \dfrac{14}{3} - \dfrac{4}{1} \cdot \dfrac{3}{3}\)



Can you do it now?
 
mmm4444bot said:
Yes -- there is a mathematical way; it's called "getting a common denominator".



We convert the fraction 4/1 to an equivalent number of thirds by multiplying 4/1 by 3/3.

\(\displaystyle \dfrac{14}{3} - \dfrac{4}{1} \cdot \dfrac{3}{3}\)



Can you do it now?
Click to expand...

\(\displaystyle \dfrac{14}{3} - \dfrac{4}{1}\cdot\dfrac{3}{3} = \dfrac{14}{3} - \dfrac{12}{3} = \dfrac{2}{3}\)

Now the problem becomes;

\(\displaystyle \dfrac{2}{3} - \dfrac{51}{3} = - \dfrac {49}{3}\)

I have got this last part by subtracting 17 from the RHS of the equals sign and done the same again as to the - 4, this ends up as shown but according to my circle drawing this coordinate seems way off the mark?
 
Probability said:
Now the problem becomes;

\(\displaystyle \dfrac{2}{3} - \dfrac{51}{3} = - \dfrac {49}{3}\)

I have got this last part by subtracting 17 from the RHS of the equals sign and done the same again as to the - 4, this ends up as shown
Click to expand...

I do not understand the statement in red.

As for the blue statement, what ends up as shown where?


After you replaced the incorrect number 10/3 with the correct number 2/3, you proceeded to square (-5x/3+2/3), yes?


(x + 3)^2 + (-5x/3 + 2/3)^2 = 17
 
mmm4444bot said:
I do not understand the statement in red.

As for the blue statement, what ends up as shown where?


After you replaced the incorrect number 10/3 with the correct number 2/3, you proceeded to square (-5x/3+2/3), yes?


(x + 3)^2 + (-5x/3 + 2/3)^2 = 17
Click to expand...

Sorry my fault for not explaining clearly. The minus 4 which you kindly explained how to convert to (2/3) I then took the = 17 and moved to the LHS and said ( - 17), then converted it as you did the (- 4) and I ended up with ( - 49/3).


I don't think I got this right or its not finished as the circle I have I am looking at drawing a line either does not intersect the circle, or is a tangent to the circle, or cuts the circle and will end up with two roots.


I seem to be losing sight of were I am going with this problem.


X^2 + 6x + 9 + ( - 5/3x - 49/3) = 0


I could factor the LHS but not sure what to do with the RHS?


I also don't think that - 49/3 is right, the position of this on the unit circle is way off and the question is saying that the line intersects, so I am definately going wrong somewhere?
 
Probability said:
x^2 + 6x + 9 + ( - 5/3x - 49/3) = 0
Click to expand...


I asked whether you squared (-5x/3 + 2/3), but you did not answer that question.

I don't understand how (-5x/3 + 2/3)^2 - 17 changed into -5x/3 - 49/3; that's not correct.

The x^2 + 6x + 9 part is correct, however.

Double check your work on simplifying the (-5x/3 + 2/3)^2 - 17 part.

If you cannot find your error(s), please post your work.



Probability said:
I seem to be losing sight of were I am going with this problem.
Click to expand...

You're trying to find the values of x where the line intersects the circle.



You correctly found the equation of the circle:

(x + 3)^2 + (y - 4)^2 = 17



You correctly substituted y = -5x/3 + 14/3:

(x + 3)^2 + (-5x/3 + 14/3 - 4)^2 = 17



You corrected your subtraction error, to get this:

(x + 3)^2 + (-5x/3 + 2/3)^2 = 17



You expanded (multiplied out) the square of x+3, to get this:

x^2 + 6x + 9 + (-5x/3 + 2/3)^2 = 17



The next step is to expand (-5x/3 + 2/3)^2

After that, subtract 17 from both sides, and finally combine all like-terms.

Your goal is to manipulate the equation into standard form ax^2+bx+c=0, so that you may then solve for x.
 
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mmm4444bot said:


I asked whether you squared (-5x/3 + 2/3), but you did not answer that question.

I don't understand how (-5x/3 + 2/3)^2 - 17 changed into -5x/3 - 49/3; that's not correct.

The x^2 + 6x + 9 part is correct, however.

Double check your work on simplifying the (-5x/3 + 2/3)^2 - 17 part.

If you cannot find your error(s), please post your work.





You're trying to find the values of x where the line intersects the circle.



You correctly found the equation of the circle:

(x + 3)^2 + (y - 4)^2 = 17



You correctly substituted y = -5x/3 + 14/3:

(x + 3)^2 + (-5x/3 + 14/3 - 4)^2 = 17



You corrected your subtraction error, to get this:

(x + 3)^2 + (-5x/3 + 2/3)^2 = 17



You expanded (multiplied out) the square of x+3, to get this:

x^2 + 6x + 9 + (-5x/3 + 2/3)^2 = 17



The next step is to expand (-5x/3 + 2/3)^2

Did I get this right;

(-5x/3 + 2/3)(-5x/3 + 2/3) = 25x/3 - 10x/3 - 10x/3 + 4/3 = 17

5x/3 + 4/3 = 17

5x/3 + 4/3 - 17/1 x 3/3 = 0

5x/3 + 4/3 - 51/3 = 0

5x/3 - 47/3 = 0

After that, subtract 17 from both sides, and finally combine all like-terms.

Your goal is to manipulate the equation into standard form ax^2+bx+c=0, so that you may then solve for x.
Click to expand...

Not sure about my methods above, but to me I still have mixed linear equations with fractions, so I still am unsure if my method is OK or I am suppose to somehow remove the fractions?
 
-5x/3 times -5x/3 is not 25x/3

You only multiplied -5 times -5. That's not good enough!

x times x is x^2, not x

3 times 3 is 9, not 3

I think that you need classroom help or face-to-face tutoring, in order to learn:

(-5x/3)(-5x/3) = 25x^2/9

I did not look at any of your other work. Here is the standard form of the equation that results from proper simplifications:

(34/9)x^2 + (34/9)x - 68/9 = 0

Solving this equation gives two solutions for x; those two values are the x-coordinates of the two intersection points for the circle and line in this exercise.
 
mmm4444bot said:
-5x/3 times -5x/3 is not 25x/3

You only multiplied -5 times -5. That's not good enough!

x times x is x^2, not x

3 times 3 is 9, not 3

I think that you need classroom help or face-to-face tutoring, in order to learn:

(-5x/3)(-5x/3) = 25x^2/9

I did not look at any of your other work. Here is the standard form of the equation that results from proper simplifications:

(34/9)x^2 + (34/9)x - 68/9 = 0

Solving this equation gives two solutions for x; those two values are the x-coordinates of the two intersection points for the circle and line in this exercise.
Click to expand...

I must admit I have no idea how you arrived at the quadratic in blue above, I am left wondering if it is indeed part of the same question?

In my question I have;

(-5x/3 + 2/3) (-5x/3 + 2/3) = 25x^2 - 20x + 4


1/9 (25x^2 - 20x + 4) = 25x^2/9 - 20x/9 + 4/9

From here I don't know if the above is correct method and I am not sure how to combine the above to;

x^2 + 6x + 9 + 25x^2/9 - 20x/9 + 4/9

What I seem to end up with is a mixed linear expression and fractions?

Which brings me back to where I was originally, and that was trying to get rid of fractions to make the expression correct?
 
Can somebody please explain to me when you are asked to multiply both sides of the expression that some parts are not actually multiplied?

Example;

x^2 + 6x + 9 + 25x^2/9 - 20x/9 + 4/9 = 17

9x^2 + 54x + 81 + 25x^2/9 - 20x/9 + 4/9 = 153


So the red section is left alone?
 
Probability said:
Can somebody please explain to me when you are asked to
multiply both sides of the equation that some parts are not actually multiplied?
The distributive rule is consistent.

Example;

x^2 + 6x + 9 + 25x^2/9 - 20x/9 + 4/9 = 17

9x^2 + 54x + 81 + 25x^2/9 - 20x/9 + 4/9 = 153 No.


So the red section is left alone? No.
Click to expand...

1) If it is desired to clear the denominators, 2) you have the correct equation,
and 3) it is a "step in the right direction" to eliminate the fractions, then look here:


\(\displaystyle x^2 \ + \ 6x \ + \ 9 \ + \ \dfrac{25x^2}{9} \ - \ \dfrac{20x}{9} \ + \ \dfrac{4}{9} \ = \ 17\)


\(\displaystyle 9\bigg(x^2 \ + 6x \ + \ 9 \ + \ \dfrac{25x^2}{9} \ - \ \dfrac{20x}{9} \ + \ \dfrac{4}{9}\bigg) \ = \ 9(17)\)


\(\displaystyle 9(x^2) \ + \ 9(6x) \ + \ 9(9) \ + \ \dfrac{9}{1}\bigg(\dfrac{25x^2}{9}\bigg) \ - \ \dfrac{9}{1}\bigg(\dfrac{20x}{9}\bigg) \ + \ \dfrac{9}{1}\bigg(\dfrac{4}{9}\bigg) \ = \ 9(17)\)


\(\displaystyle 9x^2 \ + \ 54x \ + \ 81 \ + \ 25x^2 \ - \ 20x \ + \ 4 = 153\)
 
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lookagain said:
1) If it is desired to clear the denominators, 2) you have the correct equation,
and 3) it is a "step in the right direction" to eliminate the fractions, then look here:


\(\displaystyle x^2 \ + \ 6x \ + \ 9 \ + \ \dfrac{25x^2}{9} \ - \ \dfrac{20x}{9} \ + \ \dfrac{4}{9} \ = \ 17\)


\(\displaystyle 9\bigg(x^2 \ + 6x \ + \ 9 \ + \ \dfrac{25x^2}{9} \ - \ \dfrac{20x}{9} \ + \ \dfrac{4}{9}\bigg) \ = \ 9(17)\)


\(\displaystyle 9(x^2) \ + \ 9(6x) \ + \ 9(9) \ + \ \dfrac{9}{1}\bigg(\dfrac{25x^2}{9}\bigg) \ - \ \dfrac{9}{1}\bigg(\dfrac{20x}{9}\bigg) \ + \ \dfrac{9}{1}\bigg(\dfrac{4}{9}\bigg) \ = \ 9(17)\)


\(\displaystyle 9x^2 \ + \ 54x \ + \ 81 \ + \ 25x^2 \ - \ 20x \ + \ 4 = 153\)
Click to expand...

Thank you for the effort you put in above, very much appreciated. In my previous thread I was asking why the RHS was left alone and you said no, I was referring to the RED section I marked and not after the equals sign, just in case we didn't meet on that idea.

There must be a mathematical reason why when multiplying through by 9 that the RED part is cancelled down instead of multiplying through, this was a part I was unclear about?

The other idea I am not fully clear on is that if you were to say as above, multiply through by 9 to clear the denominators, then if the RHS 17 x 9 = 153, and then the LHS starting at x^2 becomes 9x^2 + 54x + 81, then up to that point I understand, then the problem became that further along at 25x^2 this was not multiplied by 9, but the 9 was being used as a multiplier moving to the numerator position along the LHS, yet it is still in the denominator as a fraction and instead of multiplying it is now cancelled?

Now I admit that I have seen many examples of maths do this and I must admit that some areas of maths don't do this as this then leads to the incorrect solution, especially in some branches of engineering and science, so this has caused me some confusion also?

I don't know what to put it down to, either lack of experience, training or tutors and materials not explaining the course correctly I don't know?

I know the course material does not give worked examples and does not give solutions to work to for a guide, so the questions asked based on the written material you would soon loose sight of where you are suppose to be heading, some of the material is like the blind leading the blind, which is why I ask so many questions because the material is not very well written.
 
Denis said:
9x^2 + 54x + 81 + 25x^2/9 - 20x/9 + 4/9 = 153
Incorrect; should be:
9x^2 + 54x + 81 + 25x^2 - 20x + 4 = 153

Like, 20x/9 * 9 = 20x.

Perhaps you needed another coffee :rolleyes:
Click to expand...

Thanks Denis, I think I confused myself when I said 3 x 3 = 9, but then wrote 25x^2/9 - 20x/9 + 4/9.

Somehow I forgot that I had only got one 9?

So when I multiply through by the ONLY ONE 9 I got then I end up with;

9x^2 + 54x + 81 + 25x^2 - 20x + 4 = 153

I see it OK now but in the interim of confusing myself and trying to learn it
:shock:

Glad I think I understand now:smile:
 
lookagain said:
\(\displaystyle \dfrac{9}{1}\bigg(\dfrac{25x^2}{9}\bigg) \ = \ \dfrac{9(25x^2)}{9} \ = \ 25x^2.\)
Click to expand...

Probability,

what was my motivation for writing the product above this way?


I wanted it to be clear that \(\displaystyle 9\bigg(\dfrac{25x^2}{9}\bigg) \ \ne \ \dfrac{25x^2}{9(9)}, \ \ \)

or that \(\displaystyle 9\bigg(\dfrac{25x^2}{9}\bigg) \ \ne \ \dfrac{9(25x^2)}{9(9)}.\)
 
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