SOLUTION: Find the center and radius of the circle x^2+y^2-2x+8y+8=0

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Question 363220: Find the center and radius of the circle
x^2+y^2-2x+8y+8=0
Found 2 solutions by Fombitz, solver91311:
Answer by Fombitz(32388) About Me  (Show Source):
You can put this solution on YOUR website!

Complete the square in x and y,
x%5E2%2By%5E2-2x%2B8y%2B8=0
x%5E2-2x%2By%5E2%2B8y%2B8=0
%28x%5E2-2x%2B1%29%2B%28y%5E2%2B8y%2B16%29%2B8=1%2B16
%28x-1%29%5E2%2B%28y%2B4%29%5E2=17-8
%28x-1%29%5E2%2B%28y%2B4%29%5E2=9
The general equation for a circle centered at (h,k) with a radius R is:
%28x-h%29%5E2%2B%28y-k%29%5E2=R%5E2
Comparing,
(h,k)=(1,-4)
R=3

Answer by solver91311(24713) About Me  (Show Source):
You can put this solution on YOUR website!


A circle centered at with radius is represented by the equation:



For



You need to complete the square on both and and then factor the two resulting trinomials to put the equation into the above form.

Step 1:

Put the constant term into the RHS and rearrange the remaining terms in the LHS:



Step 2:

Take the coefficient on the first degree term, divide by 2, square the result and then add that result to both sides.



Step 3:

Do the same thing with the coefficient on the first degree term.



And collect the terms in the RHS



Step 4:

Factor the two perfect square trinomials:



The center and radius can now be determined by inspection.

John

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