The circle with the equation x^2+y^2-2x+4y=0

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In summary, the conversation discusses finding the appropriate place to post about geometry on the forum and solving a circle equation with given center and radius options. One method mentioned is completing the square to manipulate the equation into the standard form of a circle, while another involves using the formula r=√(A2+B2-4C)/2 to find the radius and the coordinates of the center. The formula for finding the center is derived and explained in detail.
  • #1
Jaco Viljoen
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Mod note: Moved from technical math section, so no template.
Hi All, i couldn't find any geometry under coursework and homework, but saw some geometry under general math.
I hope this is the right place, please advise and move if I have posted in the wrong place.The circle with equation x2+y2-2x+4y=0 has a centre C and radius r where
1. C(1,-2); r=Sqrt 5
2. C(-1,2); r=Sqrt 5
3. C(1,-4); r=sqrt 17
4. C(-1,4); r=17
5. C(1,-2); r=5

I have tried a couple of things:
Simplifying the equation:
x2+y2-2x+4y=0
x(x-2)+y(y+4)
x=2 and y=-4 This doesn't correspond to any of my options so can't be correct?

Using the standard form of the equation of a circle:

1.
(x-h)2+(y-k)2=r2
(x-1)2+(y+2)2=r2
x2-x-x+1+y2+2y+2y+4=r2
x2-2x+1+y2+4y+4=r
x2+y2-2x+4y+5=r2
x2+y2-2x+4y=-5
radius can't be negative...
so r=sqrt5

2.
(x-h)2+(y-k)2=r2
(x+1)2+(y-2)2=r2
(x+1)(x+1)+(y-2)(y-2)=r2
x2+1x+1x+1+y2-2y-2y-4=r2
x2+2x+y2-4y-3=r2
x2+y2+2x-4y=sqrt3
This doesn't correspond to any of my options so can't be correct.

3.
(x-h)2+(y-k)2=r2
(x-1)2+(y-(-4))2=22
(x-1)(x-1)+(y-2)(y-2)=r2
x2-1x-1x+1+y2-2y-2y+4=r
x2+y2-2x-4y+5=r2
doesn't match the circle's equation

4.
(x-h)2+(y-k)2=r2
(x+1)2+(y-4)2=r2
(x+1)(x+1)+(y-4)(y-4)=r2
x2+1x+1x+1+y2-4y-4y+8=r2
x2+y2+2x-8y+9=r2
doesn't match the circle's equation

5.
(x-h)2+(y-k)2=r2
(x-1)2+(y+2)2=r2
x2-x-x+1+y2+2y+2y+4=r2
x2-2x+1+y2+4y+4=r
x2+y2-2x+4y+5=r2

I would really appreciate some direction on this.
Thank you all.
 
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  • #2
Hello Jaco,
when you have a circle equation in form of x2+y2+Ax+By+C=0 in order to find the radius you should use the equation r=√(A2 + B2 -4C) / 2
and for the center of the circle K( -A / 2 , - B / 2 ) if you replace the variables with the numbers you should get the 1st answer.
 
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  • #3
Thank you Nick,
I do not know these formulas, or just don't recognise them.(where can I find these formulas? If you don't mind?)

Is what I have done here correct?
1.
(x-h)2+(y-k)2=r2
(x-1)2+(y+2)2=r2
x2-x-x+1+y2+2y+2y+4=r2
x2-2x+1+y2+4y+4=r
x2+y2-2x+4y+5=r2
x2+y2-2x+4y=-5
radius can't be negative...
so r=sqrt5

Thank you again.
Jaco
 
  • #4
Jaco, don't use the formulae that Nick mentioned. The answers can more easily be found by completing the square similarly to what you've been trying to do.

There are two kinds of problems of these types: either you're given the details of the circle, such as it has centre (h,k) and radius r, which then means the equation for the circle is
[tex](x-h)^2+(y-k)^2=r^2[/tex]

or you're expected to go in the opposite direction, which is to be given an equation and find the details of the circle from that. This is always the more interesting one because they'll never give you the equation in the form above, so you have to manipulate it into such a form to get the answers.

So you're given
[tex]x^2+y^2-2x+4y=0[/tex]

Looking just as the x terms, on the left hand side (LHS) we have [itex]x^2-2x[/itex]. Now, there is only one value of h in [itex](x-h)^2[/itex] that gives us that result (ignoring the constants), and that is of course h=1, giving us
[tex](x-1)^2=x^2-2x+1[/tex]
If we chose any other h then we'd have a different coefficient of x and would thus have the wrong answer. But what about the +1? Well, 1-1=0, right? So let's do just that to make the two expressions equal.

We have
[tex]x^2-2x[/tex]
but want to complete the square on it, and we know that
[tex](x-1)^2=x^2-2x+1[/tex]
and so
[tex]x^2-2x[/tex][tex]=x^2-2x+1-1[/tex][tex]=(x^2-2x+1)-1[/tex][tex]=(x-1)^2-1[/tex]

Now do the same for the y values.
 
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  • #5
A proof of the previous formulas I used :
Set K( Xo , Yo ) , a point where the center of the circle is located. Then from the classical circle equation : (X - Xo)2 + (Y - Yo)2 = r2 ((1)) if you expand this equation you will get
X2 - 2X*Xo + Xo2 + Y2 - 2Y*Yo + Yo2 = r2 ((2)) now if you set A = -2Xo and B = -2Yo and C=Xo2+Yo2-r2 from the equation ((2)) and bring r2 to the other side you will get A2+B2-4C = (-2Xo)2 + (-2Yo)2 -4(Xo2 + Yo2 - r2) = 4Xo2 + 4Yo2 -4Xo2 -4Yo2 +4r2 = 4r2 > 0 ((3)) now from equation ((3)) if you solve for the radius you will get the formula : r = √( A2+ B2 -4C ) / 2 . Now to find the coordinates of the center K, for ever circle in form of x2+ y2+ Ax + By + C = 0 <=> x2 + 2*(A/2)*x + A2/4 + y2+ 2*(B/2)*y + B2/4 = A2/4 + B2/4 - C if you factorize you will end up in the equation : ( x+A/2 )2 + ( y+B/2 )2 = (A2+B2 - 4C)/4 ((4)) from equation ((3)) you can simply observe that A2+B2 - 4C = 4r2 and if you replace it to the equation ((4)) you will see the classical circle equation. So the coordinates from equation ((4)) are K( -A/2 , -B/2 ) .

Now for x2+y2-2x+4y=0 you can solve in another way. You can add both sides with 5 and you 'll get : x2+ y2- 2x + 4y + 5 = 5 and by factorizing you will end up in the form
(x-1)2+(y+2)2 = 5 but r2=5 so r = √5, so the coordinates are K(1,-2) and r=√5 . Generally all you need is to factorize it to the form of (X - Xo)2+( Y - Yo )2 = r2 to find both radius and coordinates of the center without involving the equations i showed above.

Thanks for replying Jaco
 
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  • #6
Mentallic,
I am not following,
(x−1)2=x2−2x+1
I don't know why this was done, but would y be:
y2+4y
(y-(-2))2
(y+2)2=y2+4y+4
(y+2)2-4
 
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  • #7
Jaco if you multiply (x-1)(x-1) you will get x2-2x + 1
 
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  • #8
Nick Papachrist said:
Jaco if you multiply (x-1)(x-1) you will get x2-2x + 1
I understand this, but why have we removed the y terms? I have never seen or done this...
 
  • #9
he didnt removed it he just isolated it from the equation to show you that you need to complete the square of x2 + 2x by adding 1 to both sides same for y check my answer at the end .

Nick Papachrist said:
"Now for x2+y2-2x+4y=0 you can solve in another way. You can add both sides with 5 and you 'll get : x2+ y2- 2x + 4y + 5 = 5 and by factorizing you will end up in the form
(x-1)2+(y+2)2= 5 but r2=5 so r = √5, so the coordinates are K(1,-2) and r=√5 . Generally all you need is to factorize it to the form of (X - Xo)2+( Y - Yo )2= r2 to find both radius and coordinates of the center without involving the equations i showed above. "
 
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  • #10
Shucks, I am still feeling lost, I am going to see if I can find some more info on this and will try this again,
Thank you for the help, Nick and Mentallic.

I will be back on this thread as soon as I understand why these steps were taken.
Thank you again.

Jaco
 
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  • #11
Jaco Viljoen said:
I understand this, but why have we removed the y terms? I have never seen or done this...

I haven't, I've just focused on the x terms for the moment.
Let's focus on the LHS of the equation:

[tex]x^2+y^2-2x+4y[/tex]
[tex]=x^2-2x+y^2+4y[/tex]
[tex]=x^2-2x+1-1+y^2+4y+4-4[/tex]
[tex]=(x-1)^2-1+(y+2)^2-4[/tex]
 
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  • #12
Oh, I think I get it now.

[tex]x^2+y^2-2x+4y[/tex]
[tex]=x^2-2x+y^2+4y[/tex]
[tex]=x^2-2x+1-1+y^2+4y+4-4[/tex] , add half of x-term^2 and add half of y-term^2
[tex]=(x-1)^2-1+(y+2)^2-4[/tex]
[tex]=(x-1)^2+(y+2)^2=+1+4[/tex]
[tex]x=1, y=-2, r=√5[/tex]
 
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  • #13
Very good :)

And just one final thing. If you're unsure about what to add and subtract, complet the square first and then find what the constant is from that. For example, given

[tex]x^2+10x[/tex]

we know that to complete the square, we take half of the coefficient of x (5 in this case) and plug it in for h in (x+h)2, giving us
[tex](x+5)^2[/tex]
and then if we expand this expression, the constant turns out to be 25, hence we have to add and subtract 25.
 
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  • #14
Just focus on the following part of your post:
Jaco Viljoen said:
...
[tex]=(x-1)^2-1+(y+2)^2-4[/tex]
[tex]\color{red}{=}(x-1)^2+(y+2)^2=+1+4[/tex]
That leading "=" sign should not be in the last line.
##\displaystyle\ (x-1)^2-1+(y+2)^2-4 \ne (x-1)^2+(y+2)^2\ ##
 
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  • #15
Thank you for pointing this out Sammy,
 

Related to The circle with the equation x^2+y^2-2x+4y=0

1. What is the center and radius of the circle with the equation x^2+y^2-2x+4y=0?

The center and radius of a circle can be determined by rearranging the equation into the standard form (x-h)^2 + (y-k)^2 = r^2, where (h,k) is the center and r is the radius. In this case, the equation can be rewritten as (x-1)^2 + (y+2)^2 = 5, so the center is at (1,-2) and the radius is √5.

2. How do you graph the circle with the equation x^2+y^2-2x+4y=0?

To graph a circle, plot the center point and then use the radius to draw the circle around the center. In this case, the center is at (1,-2) and the radius is √5. So, plot the point (1,-2) and then draw a circle with a radius of √5 around that point.

3. What is the method for finding the intercepts of the circle with the equation x^2+y^2-2x+4y=0?

To find the x-intercepts, set y=0 and solve for x. To find the y-intercepts, set x=0 and solve for y. In this case, setting y=0 gives x^2-2x=0, which can be factored to get x=0 and x=2 as the x-intercepts. Setting x=0 gives y^2+4y=0, which can be factored to get y=0 and y=-4 as the y-intercepts.

4. Can the circle with the equation x^2+y^2-2x+4y=0 be rewritten in polar form?

Yes, any circle can be expressed in polar form as r = a cos(θ-h) + b sin(θ-k), where (h,k) is the center and a is the radius. In this case, the equation can be rewritten as r = √5 cos(θ-π/2) + √5 sin(θ-π/2).

5. How does the circle with the equation x^2+y^2-2x+4y=0 relate to the Pythagorean theorem?

The equation of a circle with a center at (h,k) and a radius of r can be rewritten as (x-h)^2 + (y-k)^2 = r^2. This is similar to the Pythagorean theorem, which states that in a right triangle, the sum of the squares of the two shorter sides is equal to the square of the hypotenuse. In this case, the circle can be visualized as a right triangle with sides of length (x-h), (y-k), and r.

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