SOLUTION: Find the equation, in standard form, of the circle. Diameter with endpoints (-2,-1) and (6,-3)

Algebra ->  Equations -> SOLUTION: Find the equation, in standard form, of the circle. Diameter with endpoints (-2,-1) and (6,-3)      Log On


   

Question 1183676: Find the equation, in standard form, of the circle.
Diameter with endpoints (-2,-1) and (6,-3)
Found 3 solutions by josgarithmetic, ikleyn, MathTherapy:
Answer by josgarithmetic(39617) About Me  (Show Source):
You can put this solution on YOUR website!
Diameter with endpoints (p,v) and (q,w)
--------
--------

Center point for the circle is some (h,k) and h=%28p%2Bq%29%2F2, and k=%28v%2Bw%29%2F2.

Radius squared is %28%28q-p%29%2F2%29%5E2%2B%28%28w-v%29%2F2%29%5E2.

Equation would be %28x-%28p%2Bq%29%2F2%29%5E2%2B%28y-%28v%2Bw%29%2F2%29%5E2=%28q-p%29%5E2%2B%28w-v%29%5E2.
Using the actual values as you were given may be a little easier.

Your circle's center is ( (-2+6)/2, (-1-3)/2 )
or (2,-2).
.
.
highlight%28%28x-2%29%5E2%2B%28y%2B2%29%5E2=cross%2868%29%29

Answer by ikleyn(52781) About Me  (Show Source):
You can put this solution on YOUR website!
.
Find the equation, in standard form, of the circle.
Diameter with endpoints (-2,-1) and (6,-3)
~~~~~~~~~~~~~~~~~ 

The center's  x-coordinate is  %28-2+%2B+6%29%2F2 = 4%2F2 = 2.

The center's  y-coordinate is  %28-1+%2B+%28-3%29%29%2F2 = -4%2F2 = -2.


The square of the diameter's length is  d^2 = ((6-(-2))^2 + ((-3)-(-1))^2 = (8^2 + 2^2) = 68.


The square of the radius is  r^2 = 68/2^2 = 68/4 = 17.


The standard form equation of the circle is


    %28x-2%29%5E2 + %28y%2B2%29%5E2 = 17.

Solved.


//////////////


The post by @josgarithmetic is  TOTALLY  WRONG.


IGNORE  it,  for the sake of your safety.



Answer by MathTherapy(10552) About Me  (Show Source):
You can put this solution on YOUR website!

Find the equation, in standard form, of the circle.
Diameter with endpoints (-2,-1) and (6,-3)
If you want to confuse someone or yourself to death, then use the many variables that the other person presented to you. 

If not, then just use the MIDPOINT formula: , to get:

                                            matrix%281%2C5%2C+%22%282%2C%22%2C+%22-+2%29%2C%22%2C+or%2C+%22%28h%2C%22%2C+%22k%29%22%29 <===== CENTER coordinates of circle

Standard form of the equation of a CIRCLE: matrix%281%2C3%2C+%28x+-+h%29%5E2+%2B+%28y+-+k%29%5E2%2C+%22=%22%2C+r%5E2%29

                                           matrix%281%2C3%2C+%28-+2+-+2%29%5E2+%2B+%28-+1+-+-+2%29%5E2%2C+%22=%22%2C+r%5E2%29 ------ In the ABOVE, substituting (2, - 2) for (h, k) and (- 2, - 1) for (x, y). NOTE that (6, - 3) could also be used for (x, y)

                                           

For the equation in Standard form, we finally get: