SOLUTION: Solve the following systems of two circles by finding the intersection point(s), if any, algebraically. x^2 + y^2 = 9 (x-3)^2 + (y+3)^2 =9

Algebra ->  Systems-of-equations -> SOLUTION: Solve the following systems of two circles by finding the intersection point(s), if any, algebraically. x^2 + y^2 = 9 (x-3)^2 + (y+3)^2 =9      Log On


   

Question 957021: Solve the following systems of two circles by finding the intersection point(s), if any, algebraically.
x^2 + y^2 = 9
(x-3)^2 + (y+3)^2 =9
Answer by Fombitz(32388) About Me  (Show Source):
You can put this solution on YOUR website!
%28x-3%29%5E2+=9-+%28y%2B3%29%5E2
x-3=0+%2B-+sqrt%289-+%28y%2B3%29%5E2%29
x=3+%2B-+sqrt%289-+%28y%2B3%29%5E2%29
So then
x%5E2=9%2B6sqrt%289-%28y%2B3%29%5E2%29%2B9-%28y%2B3%29%5E2-Positive
x%5E2=9-6sqrt%289-%28y%2B3%29%5E2%29%2B9-%28y%2B3%29%5E2-Negative
Substituting,
%289%2B6sqrt%289-%28y%2B3%29%5E2%29%2B9-%28y%2B3%29%5E2%29+%2B+y%5E2+=+9-Positive
%289-6sqrt%289-%28y%2B3%29%5E2%29%2B9-%28y%2B3%29%5E2%29+%2B+y%5E2+=+9-Negative
.
.
6sqrt%289-%28y%2B3%29%5E2%29%2B9-y%5E2-6y-9+%2B+y%5E2+=+0-Positive
-6sqrt%289-%28y%2B3%29%5E2%29%2B9-y%5E2-6y-9+%2B+y%5E2+=+0-Negative
.
.
sqrt%289-%28y%2B3%29%5E2%29=y++-Positive
-sqrt%289-%28y%2B3%29%5E2%29=y++-Negative
.
.
9-%28y%2B3%29%5E2=y%5E2
9-y%5E2-6y-9=y%5E2
2y%5E2%2B6y=0
y%28y%2B3%29=0
Two solutions:
y=0
Then,
%28x-3%29%5E2+%2B+%280%2B3%29%5E2+=9
%28x-3%29%5E2=0
x=3
and
y%2B3=0
y=-3
Then,
x%5E2%2B%28-3%29%5E2=9
x=0
(3,0) and (0,-3)
.
.
.