Equation of a circle...

[shawie]

how do i find the answer to this?

The center is (2,3); the circle passses through (5,6)

wouldn't it be something like... (x-2)^2+(y-3)^2=.... yeah i don't know, help please?

oh and there's another question that i just don't know how to approach:

Determine whether the line 3x+2y=6 contains the center of the circle x^2+y2+4x-12y+24=0.

 

[Ted]

You're very close on the first part. You figured out correctly how to center the circle at (2,3). The part you need is the radius. Now you have to stop and think... you know the center and you know a point on the circle. The radius is the distance between the two points. Knowing that, can you figure out what the radius is and what the circle equation should be?

For the second part you want to factor the equation so that it's in the typical form like you used in the first part. Then you can figure out what its center point is. Finally, just see if that point works when you plug it into the line.

Does this help you get on track?

Ted

 

[shawie]

You're very close on the first part. You figured out correctly how to center the circle at (2,3). The part you need is the radius. Now you have to stop and think... you know the center and you know a point on the circle. The radius is the distance between the two points. Knowing that, can you figure out what the radius is and what the circle equation should be?

...i don't know...

For the second part you want to factor the equation so that it's in the typical form like you used in the first part. Then you can figure out what its center point is. Finally, just see if that point works when you plug it into the line.

...erm, i tried something.... am i off?

 

[galactus]

You're very close on the first part. You figured out correctly how to center the circle at (2,3). The part you need is the radius. Now you have to stop and think... you know the center and you know a point on the circle. The radius is the distance between the two points. Knowing that, can you figure out what the radius is and what the circle equation should be?

...i don't know...

For the second part you want to factor the equation so that it's in the typical form like you used in the first part. Then you can figure out what its center point is. Finally, just see if that point works when you plug it into the line.

...erm, i tried something.... am i off?

You just have your signs backwards.

\(\displaystyle (x+2)^{2}+(y-6)^{2}=16\)

You can easily see what the center coordinates are from this equation. Enter them into the line equation you were given and see if it works.

 

[Unco]

You're very close on the first part. You figured out correctly how to center the circle at (2,3). The part you need is the radius. Now you have to stop and think... you know the center and you know a point on the circle. The radius is the distance between the two points. Knowing that, can you figure out what the radius is and what the circle equation should be?

...i don't know...

Ted gave an excellent explanation.

How about a sketch:

                         /|\ y
                          |               (5,6)
                         6+                 x  <- pt on circle
                          |               / 
                          |             / _
                          |           /  |\ 
                          |   (2,3) /      \  
                         3+       x          this length >must< be the radius   
                          |    center
                          |
                          |                     
                          |
                 ---------+-------+---------+---->
                          |       2         5    x
                          |
                          |
                          |

The distance between two points is Pythagoras:

. . \(\displaystyle \L \, r \, = \sqrt{(2 \, - 5)^2 \, + \, (3 \, - \, 6)^2 \, \, }\)

 

[shawie]

You're very close on the first part. You figured out correctly how to center the circle at (2,3). The part you need is the radius. Now you have to stop and think... you know the center and you know a point on the circle. The radius is the distance between the two points. Knowing that, can you figure out what the radius is and what the circle equation should be?

...i don't know...

For the second part you want to factor the equation so that it's in the typical form like you used in the first part. Then you can figure out what its center point is. Finally, just see if that point works when you plug it into the line.

...erm, i tried something.... am i off?

You just have your signs backwards.

\(\displaystyle (x+2)^{2}+(y-6)^{2}=16\)

You can easily see what the center coordinates are from this equation. Enter them into the line equation you were given and see if it works.

what do you mean by that?

 

[shawie]

You're very close on the first part. You figured out correctly how to center the circle at (2,3). The part you need is the radius. Now you have to stop and think... you know the center and you know a point on the circle. The radius is the distance between the two points. Knowing that, can you figure out what the radius is and what the circle equation should be?

...i don't know...

Ted gave an excellent explanation.

How about a sketch:

                         /|\ y
                          |               (5,6)
                         6+                 x  <- pt on circle
                          |               / 
                          |             / _
                          |           /  |\ 
                          |   (2,3) /      \  
                         3+       x          this length >must< be the radius   
                          |    center
                          |
                          |                     
                          |
                 ---------+-------+---------+---->
                          |       2         5    x
                          |
                          |
                          |

The distance between two points is Pythagoras:

. . \(\displaystyle \L \, r \, = \sqrt{(2 \, - 5)^2 \, + \, (3 \, - \, 6)^2 \, \, }\)

ohhh ok, wow i get it! thank you guys so much!

 

[stapel]

what do you mean by that?

What does who mean by which?

Eliz.

 

[shawie]

...

i quoted galactus, i don't know what they meant by:

Enter them into the line equation you were given and see if it works.

 

[galactus]

All I meant was use the line equation with your circle centers.

The line equation is 3x+2y=6, isn't it?.

Your center is x=-2 and y=6

3(-2)+2(6)=6, Check!...It contains the center.

 

[shawie]

ohhhh ok, that's what you meant! alright, thanks!