Finding equation of circle.

May 2010
18
0
Ok so it says on the question:

Write the equation of the following circle:
with diameter AB and given A(4,3) and B(6,-1)

please need help really fast (Crying)(Crying)
 

dwsmith

MHF Hall of Honor
Mar 2010
3,093
582
Florida
Ok so it says on the question:

Write the equation of the following circle:
with diameter AB and given A(4,3) and B(6,-1)

please need help really fast (Crying)(Crying)
The mid point is the center of the circle

\(\displaystyle (x-x_0)^2+(y-y_0)^2=r^2\)
 

Prove It

MHF Helper
Aug 2008
12,883
4,999
Ok so it says on the question:

Write the equation of the following circle:
with diameter AB and given A(4,3) and B(6,-1)

please need help really fast (Crying)(Crying)
Find the distance between these two points using Pythagoras' Theorem. Halve it and you have the radius (\(\displaystyle r\)).

Find the midpoint of the segment between these two points. This is your centre \(\displaystyle (h, k)\).


The formula of the circle is

\(\displaystyle (x - h)^2 + (y - k)^2 = r^2\).
 
May 2010
18
0
Find the distance between these two points using Pythagoras' Theorem. Halve it and you have the radius (\(\displaystyle r\)).

Find the midpoint of the segment between these two points. This is your centre \(\displaystyle (h, k)\).


The formula of the circle is

\(\displaystyle (x - h)^2 + (y - k)^2 = r^2\).

ok so what i do is i take the points and put it on an A^2+B^2=C^2?


coz i dont really get pythagoras on this i dont see ABC(Worried)
 

dwsmith

MHF Hall of Honor
Mar 2010
3,093
582
Florida
Once you have the mid point, you can figure out the distance to one of the points using the distance formula. The distance you will have solved for will be the radius.
 

Prove It

MHF Helper
Aug 2008
12,883
4,999
ok so what i do is i take the points and put it on an A^2+B^2=C^2?


coz i dont really get pythagoras on this i dont see ABC(Worried)
The distance formula

\(\displaystyle D = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}\)

is a direct application of Pythagoras' Theorem.
 
May 2010
18
0
Once you have the mid point, you can figure out the distance to one of the points using the distance formula. The distance you will have solved for will be the radius.
uh can you show me step by step (Crying)
like plug in the given then ima solve the other stuff(Nerd)
 
May 2010
18
0
The distance formula

\(\displaystyle D = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}\)

is a direct application of Pythagoras' Theorem.

oh crap i used that and accidentally exchanged x_2 with x_1 and also for the y

(Headbang)thats why i keep getting this square root of 20 =.=
 

Prove It

MHF Helper
Aug 2008
12,883
4,999
You need to use:


  • The Distance Formula
\(\displaystyle D = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}\)

which is a direct application of Pythagoras' Theorem.


  • The Midpoint Formula
\(\displaystyle (x_m, y_m) = \left(\frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2}\right)\).
 

Prove It

MHF Helper
Aug 2008
12,883
4,999
oh crap i used that and accidentally exchanged x_2 with x_1 and also for the y

(Headbang)thats why i keep getting this square root of 20 =.=
\(\displaystyle \sqrt{20} = 2\sqrt{5}\) is correct.
 
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