SOLUTION: which of the following points is not on the circle represented by the equation {{{(x-3)^2+(y+4)^2=25}}}? A. (6,0) B. (-1,-1) C. (4,1) D. None of these is on the circle

Algebra ->  Rectangles -> SOLUTION: which of the following points is not on the circle represented by the equation {{{(x-3)^2+(y+4)^2=25}}}? A. (6,0) B. (-1,-1) C. (4,1) D. None of these is on the circle      Log On


   

Question 375357: which of the following points is not on the circle represented by the equation %28x-3%29%5E2%2B%28y%2B4%29%5E2=25?
A. (6,0)
B. (-1,-1)
C. (4,1)
D. None of these is on the circle
Found 2 solutions by Fombitz, EdwinMcCravy:
Answer by Fombitz(32388) About Me  (Show Source):
You can put this solution on YOUR website!
Plug each point in the equation.
I'll do the first, you do the rest.
%28x-3%29%5E2%2B%28y%2B4%29%5E2=25
%286-3%29%5E2%2B%280%2B4%29%5E2=25
3%5E2%2B4%5E2=25
9%2B16=25
25=25
True, so (6,0) is on the circle.
Now you also know D) is not the solution.
Now check the remaining two points.

Answer by EdwinMcCravy(4) About Me  (Show Source):
You can put this solution on YOUR website!
which of the following points is not on the circle represented by the equation %28x-3%29%5E2%2B%28y%2B4%29%5E2=25?
A. (6,0)
Let's find out if (6,0) is or is not on the circle. To do that sustitute
6 for x and 0 for y in the equation for the circle:
%28x-3%29%5E2%2B%28y%2B4%29%5E2=25
%286-3%29%5E2%2B%280%2B4%29%5E2=25
%283%29%5E2%2B%284%29%5E2=25
9%2B16=25
25=25
That is true so therefore (6,0) is a point on the circle with the
equation %28x-3%29%5E2%2B%28y%2B4%29%5E2=25
-----------------------
B. (-1,-1)
Let's find out if (-1,-1) is or is not on the circle. To do that sustitute
-1 for x and -1 for y in the equation for the circle:
%28x-3%29%5E2%2B%28y%2B4%29%5E2=25
%28-1-3%29%5E2%2B%28-1%2B4%29%5E2=25
%28-4%29%5E2%2B%283%29%5E2=25
16%2B9=25
25=25
That is true so therefore (-1,-1) is a point on the circle with the
equation %28x-3%29%5E2%2B%28y%2B4%29%5E2=25
-----------------------
C. (4,1)
Let's find out if (4,1) is or is not on the circle. To do that sustitute
4 for x and 1 for y in the equation for the circle:
%28x-3%29%5E2%2B%28y%2B4%29%5E2=25
%284-3%29%5E2%2B%281%2B4%29%5E2=25
%281%29%5E2%2B%285%29%5E2=25
1%2B25=25
26=25
That is false so therefore (4,1) is NOT a point on the circle with the
equation %28x-3%29%5E2%2B%28y%2B4%29%5E2=25
Below is an accurate graph of the circle and you can see that (4,1)
is just above the circle whereas the other two points are right on
the circle.



Edwin