finding radius and center of an equation

[abel muroi]

I was given a problem that has the equation (x - 3)^2 + y^2 = 16

I was also asked to find the center and radius of this equation, but i am having some trouble finding the center.

I assume the center is (3, 1) and the radius is 4.

is this correct?

 

[Ishuda]

I was given a problem that has the equation (x - 3)^2 + y^2 = 16

I was also asked to find the center and radius of this equation, but i am having some trouble finding the center.

I assume the center is (3, 1) and the radius is 4.

is this correct?

Not quite but almost. The equation for a circle can be written as
$\displaystyle (x - x_0)^2 + (y - y_0)^2 = r^2$
where $\displaystyle (x_0, y_0)$ is the center of the circle and r is the radius.

So, if
$\displaystyle (y - y_0)^2 = y^2$
what is $\displaystyle y_0$?

 

[abel muroi]

Not quite but almost. The equation for a circle can be written as
$\displaystyle (x - x_0)^2 + (y - y_0)^2 = r^2$
where $\displaystyle (x_0, y_0)$ is the center of the circle and r is the radius.

So, if
$\displaystyle (y - y_0)^2 = y^2$
what is $\displaystyle y_0$?

then the center must be (3,0)? since we are multiplying 0 and 0 right?

y_0 doesn't have a value, so I assumed it is 0.

 

[Steven G]

then the center must be (3,0)? since we are multiplying 0 and 0 right?

y_0 doesn't have a value, so I assumed it is 0.

The center is (3,0) but not because of any multiplication.
The center is (3,0) because of what Ishuda said, namely the equation for a circle can be written as
[FONT=MathJax_Main]([/FONT][FONT=MathJax_Math]x[/FONT][FONT=MathJax_Main]−[/FONT][FONT=MathJax_Math]x[/FONT][FONT=MathJax_Main]0[/FONT][FONT=MathJax_Main])[/FONT][FONT=MathJax_Main]2[/FONT][FONT=MathJax_Main]+[/FONT][FONT=MathJax_Main]([/FONT][FONT=MathJax_Math]y[/FONT][FONT=MathJax_Main]−[/FONT][FONT=MathJax_Math]y[/FONT][FONT=MathJax_Main]0[/FONT][FONT=MathJax_Main])[/FONT][FONT=MathJax_Main]2[/FONT][FONT=MathJax_Main]=[/FONT][FONT=MathJax_Math]r[/FONT][FONT=MathJax_Main]2[/FONT]
where [FONT=MathJax_Main]([/FONT][FONT=MathJax_Math]x[/FONT][FONT=MathJax_Main]0[/FONT][FONT=MathJax_Main],[/FONT][FONT=MathJax_Math]y[/FONT][FONT=MathJax_Main]0[/FONT][FONT=MathJax_Main])[/FONT] is the center of the circle and r is the radius.

Basically you set what is being squared to 0 and you get the x and y values of the center.

(x-3)^2 so we set x-3=0 and get x=3
y^2 so we set y=0 and well we get y=0
So the center is (x,y)=(3,0)

 

[abel muroi]

Thank you I understand now. But I have one more question.



I am asked to find the radius and center of this equation 2x² + 2y² - 3x = 0

What should I do with the -3x? can I just subtract it from 2x² ?

if that's true then..

2x² + 2y² - 3x = 0
-1x² + 2y² = 0
(x + 1)² + (y - 2)² =0

center = (-1, 2)
radius = 0

Did I make a mistake?

 

[Steven G]

Thank you I understand now. But I have one more question.



I am asked to find the radius and center of this equation 2x² + 2y² - 3x = 0

What should I do with the -3x? can I just subtract it from 2x² ?

if that's true then..

2x² + 2y² - 3x = 0
-1x² + 2y² = 0
(x + 1)² + (y - 2)² =0

center = (-1, 2)
radius = 0

Did I make a mistake?

Suppose you have 7 two's and subtract 3 two's, then you have 4 two's or 8.
If you have 8 three's and subtract 11 nine's you do not have -3 of anythings (I got -3 from 8-11).

Usually when you square a number you get a different number. So just like when you have 8 three's and subtract 3 nine's you do not have 5 anythings if you have 2x^2 and subtract 3x you do not have -1 of anything.

A circle can be written as (x-a)^2 + (y-b)^2 = r*2 where the center is at (a,b) and the radius is r. If the equation is written as c(x-a)^2 + c(y-b)^2 = r*2, then divide by c (c must be positive, why??).

I would divide both sides of 2x² + 2y² - 3x = 0 by 2 to get [x^2-(3/2)x ]+ y^2 =0

You need to complete the square for x^2-(3/2)x by taking (1/2) of [-3/2] and squaring it. Then add it to both sides of the equation.

[(1/2)(-3/2)]^2 = [--3/4]^2 = 9/16

[x^2-(3/2)x + 9/16]+ y^2 =0+ 9/16.

What is in [] in the line above is a perfect square. It is (x-3/4)^2

So the equation in standard form is (x-3/4)^2 + y^2 =9/16.

Now you can read off the center and radius. What are they??

 

[Ishuda]

then the center must be (3,0)? since we are multiplying 0 and 0 right?

y_0 doesn't have a value, so I assumed it is 0.

y₀ does have a value and you are correct when you say it is zero and that the center is (3,0).

 

[abel muroi]

Suppose you have 7 two's and subtract 3 two's, then you have 4 two's or 8.
If you have 8 three's and subtract 11 nine's you do not have -3 of anythings (I got -3 from 8-11).

Usually when you square a number you get a different number. So just like when you have 8 three's and subtract 3 nine's you do not have 5 anythings if you have 2x^2 and subtract 3x you do not have -1 of anything.

A circle can be written as (x-a)^2 + (y-b)^2 = r*2 where the center is at (a,b) and the radius is r. If the equation is written as c(x-a)^2 + c(y-b)^2 = r*2, then divide by c (c must be positive, why??).

I would divide both sides of 2x² + 2y² - 3x = 0 by 2 to get [x^2-(3/2)x ]+ y^2 =0

You need to complete the square for x^2-(3/2)x by taking (1/2) of [-3/2] and squaring it. Then add it to both sides of the equation.

[(1/2)(-3/2)]^2 = [--3/4]^2 = 9/16

[x^2-(3/2)x + 9/16]+ y^2 =0+ 9/16.

What is in [] in the line above is a perfect square. It is (x-3/4)^2

So the equation in standard form is (x-3/4)^2 + y^2 =9/16.

Now you can read off the center and radius. What are they??

I appreciate the work you did in your comment but I would prefer if you can just give me a few hints instead of giving me the answer

but thank you very much! so the center is (3/4, 0) and the radius = 3/4 after square rooting it

 

[Steven G]

I appreciate the work you did in your comment but I would prefer if you can just give me a few hints instead of giving me the answer

but thank you very much! so the center is (3/4, 0) and the radius = 3/4 after square rooting it

Fair enough regarding your request. You seem like a dedicated student so I gave you a complete solution so you can do another similar problem.
Please do me a favor and do another similar problem so that you do one on your own.

Actually here is a problem. Find the center and radius for 3x^2 + 3y^2 + 15y = 0.

Thanks!

 

[abel muroi]

Fair enough regarding your request. You seem like a dedicated student so I gave you a complete solution so you can do another similar problem.
Please do me a favor and do another similar problem so that you do one on your own.

Actually here is a problem. Find the center and radius for 3x^2 + 3y^2 + 15y = 0.

Thanks!

Alright I'll do the problem you posted

Ok so i used your method and divided all sides by 3 and got..

3x² + 3y² + 15y = 0
x² + y² + 5y = 0
x² +(y² + 5y) = 0
x² +(y² + 1/4) = 1/4
x² + (y - 1/2)² = 1/4

so the center is (0, 1/2) and radius is 1/2 after square rooting 1/4

did i do the problem correctly?

 

[Steven G]

Alright I'll do the problem you posted

Ok so i used your method and divided all sides by 3 and got..

3x² + 3y² + 15y = 0
x² + y² + 5y = 0
x² +(y² + 5y) = 0
x² +(y² + 1/4) = 1/4
x² + (y - 1/2)² = 1/4

so the center is (0, 1/2) and radius is 1/2 after square rooting 1/4

did i do the problem correctly?

Hi, sorry but this is not correct. the problem is that you are saying that y^2 + 5y = y^2 + 1/4. That means that 5y = 1/4
You also wrote that (y² + 1/4) = (y - 1/2)² . But (y - 1/2)² = y^2 -y+1/4 not (y² + 1/4).

Read up on how to complete the square and/or look at my example which hopefully will help.

Also note the following. (x+a)^2 = x^2 +2ax + a^2. Notice a few things. (1) that the coefficient of x^2 is 1. (2). If you take 1/2 of the coefficient of the x term (2a) and (3) square that result, (1/2)(2a)=a, you will get a^2 which is the constant. If all 3 conditions have been met then you have a perfect square, namely (x+a)^2.

Example: what do you add to x^2 +6x to get a perfect square and what is that perfect square? [(1/2)(+6)]^2 = [+3]^2 = +9. So we get (x+3)^2
example: what do you add to x^2 -6x to get a perfect square and what is that perfect square? [(1/2)(-6)]^2 = [-3]^2 = +9. So we get (x-3)^2

 

[abel muroi]

Hi, sorry but this is not correct. the problem is that you are saying that y^2 + 5y = y^2 + 1/4. That means that 5y = 1/4
You also wrote that (y² + 1/4) = (y - 1/2)² . But (y - 1/2)² = y^2 -y+1/4 not (y² + 1/4).

Read up on how to complete the square and/or look at my example which hopefully will help.

Also note the following. (x+a)^2 = x^2 +2ax + a^2. Notice a few things. (1) that the coefficient of x^2 is 1. (2). If you take 1/2 of the coefficient of the x term (2a) and (3) square that result, (1/2)(2a)=a, you will get a^2 which is the constant. If all 3 conditions have been met then you have a perfect square, namely (x+a)^2.

Example: what do you add to x^2 +6x to get a perfect square and what is that perfect square? [(1/2)(+6)]^2 = [+3]^2 = +9. So we get (x+3)^2
example: what do you add to x^2 -6x to get a perfect square and what is that perfect square? [(1/2)(-6)]^2 = [-3]^2 = +9. So we get (x-3)^2

I'm not sure if I understood what you wrote (but i understood your examples and the other post you wrote). But I am guessing that I made a mistake when I was multiplying 1/2 and 5 which got me 5/2 and then i squared it which got me 25/4 then simplifying it which got me 1/4. Was that the mistake?

 

[Steven G]

I'm not sure if I understood what you wrote. But I am guessing that I made a mistake when I was multiplying 1/2 and 5 which got me 5/2 and then i squared it which got me 25/4 then simplifying it which got me 1/4. Was that the mistake?

Yep. Do you really believe that 25/4 is 1/4?

if you divide two different numbers by 4 do you really expect to get the same result?

25/4 is over 6 (it's 6 1/4) while 1/4 is less then 1.

I just want to give as many ways to show that 25/4 and 1/4 are different as I can hoping that you'll see your mistake. If 25/4 simplifies to 1/4 then these numbers are the same and naturally their difference would be 0. Is 25/4 - 1/4 = 0? Or is it that 25/4 - 25/4 = 0 and 1/4 - 1/4 = 0.

75/50 simplifies to 3/2. Are there two numbers equal to one another? Does 75/50 - 3/2 = 0?

Work on these two questions first and get back with the answers please.

 

[abel muroi]

Yep. Do you really believe that 25/4 is 1/4?

if you divide two different numbers by 4 do you really expect to get the same result?

25/4 is over 6 (it's 6 1/4) while 1/4 is less then 1.

I just want to give as many ways to show that 25/4 and 1/4 are different as I can hoping that you'll see your mistake. If 25/4 simplifies to 1/4 then these numbers are the same and naturally their difference would be 0. Is 25/4 - 1/4 = 0? Or is it that 25/4 - 25/4 = 0 and 1/4 - 1/4 = 0.

75/50 simplifies to 3/2. Are there two numbers equal to one another? Does 75/50 - 3/2 = 0?

Work on these two questions first and get back with the answers please.

OK I'll try the problem again but this time I'll post the steps that I took to solve it, so that you can give me precise feedback. Oh and i decided to not simplify 25/4 because i don't think its possible

(and yes I am still having some trouble with fractions:| )

step 1 : 3x² + 3y² + 15y = 0
step 2 : x² + y² +5y = 0 ((I divided all sides by 3))
step 3 : (x + x₀)² + (y² + 5y + 25/4) = 0 + 25/4 ((I multiplied 5 by 1/2 which got me 5/2, then I squared it by 2 which got me 25/4))
step 4 : (x + x₀)² + (y + 5/2)² =25/4 ((i square rooted 25/4 which got me 5/2)
step 5 : (x + x₀)² + (y + 5/2)² = 5/2² ((then i square rooted 25/4 again which got me 5/2)

Can you tell me in which step i made a mistake?

 

[Steven G]

OK I'll try the problem again but this time I'll post the steps that I took to solve it, so that you can give me precise feedback. Oh and i decided to not simplify 25/4 because i don't think its possible

(and yes I am still having some trouble with fractions:| )

step 1 : 3x² + 3y² + 15y = 0
step 2 : x² + y² +5y = 0 ((I divided all sides by 3))
step 3 : (x + x₀)² + (y² + 5y + 25/4) = 0 + 25/4 ((I multiplied 5 by 1/2 which got me 5/2, then I squared it by 2 which got me 25/4))
step 4 : (x + x₀)² + (y + 5/2)² =25/4 ((i square rooted 25/4 which got me 5/2) No! This time you got away with it. sqrt(25/4) is not 5/2. It is +/- (5/2). So which one do you use? Look at my previous work which is color coded on purpose. [(1/2)(+5)]^2 = [+5/2]^2 = 25/4. The sign in the red brackets is VERY important as what is in the brackets is EXACTLY what goes next to the x. That is (x +5/2)^2. If what was in the red brackets above was -2/3, then we would have (x-2/3)^2. Is this clear??
step 5 : (x + x₀)² + (y + 5/2)² = 5/2² ((then i square rooted 25/4 again which got me 5/2)

Can you tell me in which step i made a mistake?

All in all not bad at all. You biggest mistake is that you never answered the question!!! What is the center and radius of this circle?!!

You wrote 5/2² which equals 5/4 NOT 25/4. You meant to write (5/2)^2

You keep writing (x + x₀)² for x^2. Does x + x₀ =x? Either write x^2 or if you prefer write (x-0)^2 or (x+0)^2

The equation is x² + (y + 5/2)² =25/4 and the center is ____ and the radius is ____

Much better work!