SOLUTION: Find the standard form of the equation of the circle having the following​ properties: Center at the origin Containing the Point(2,-1) Type the standard form of the equation

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Question 1200235: Find the standard form of the equation of the circle having the following​ properties:
Center at the origin
Containing the Point(2,-1)
Type the standard form of the equation of this circle.
Found 3 solutions by ikleyn, josgarithmetic, math_tutor2020:
Answer by ikleyn(52893) About Me  (Show Source):
You can put this solution on YOUR website!
.

If the center of a circle is at the origin, then the standard form equation of
the circle is 

    x%5E2 + y%5E2 = r%5E2,


where "r" is the radius of the circle.


Since the point (2,-1) lies on the circle, then the radius of the circle is 
the distance to the point (2,-1) from the origin.


Therefore, using the formula for the square of the distance, we have

    r%5E2 = 2%5E2 + %28-1%29%5E2 = 4 + 1 = 5.


Taking everything together, we have the standard form of the equation 
of the circle in form

    x%5E2 + y%5E2 = 5.    ANSWER

Solved, with complete explanation.



Answer by josgarithmetic(39630) About Me  (Show Source):
You can put this solution on YOUR website!
Basic representation, %28x-h%29%5E2%2B%28y-k%29%5E2=r%5E2
Your given center and point, x%5E2%2By%5E2=2%5E2%2B%28-1%29%5E2
.
.

Answer by math_tutor2020(3817) About Me  (Show Source):
You can put this solution on YOUR website!

Circle template: %28x-h%29%5E2%2B%28y-k%29%5E2+=+r%5E2
center = (h,k)
radius = r

Since the center is the origin, we have (h,k) = (0,0)
%28x-h%29%5E2%2B%28y-k%29%5E2+=+r%5E2
updates to
%28x-0%29%5E2%2B%28y-0%29%5E2+=+r%5E2
which is simply
x%5E2%2By%5E2+=+r%5E2

Now we plug the coordinates of (x,y) = (2,-1) into that equation
x%5E2%2By%5E2+=+r%5E2
2%5E2%2B%28-1%29%5E2+=+r%5E2
4%2B1+=+r%5E2
r%5E2+=+5
We don't need to find the value of r, since we can stop at r^2

The equation
x%5E2%2By%5E2+=+r%5E2
updates to
x%5E2%2By%5E2+=+5
which is the final answer.