Find the intersection of x^2+y^2=16 and y=x+10?

Question:how is this done?? please some one help!!

Answers:
They don't intersect. The first equation is for a circle centered at (0,0): the second is a line w/ y-int of 10, and an x-int of -10.

They NEVER touch.

by the way, the answer CANNOT be (3,1) as the person above me has stated: Just try plugging it into both equations:

x^2+y^2=16
3^2+1^2=16
9+1=16
10=16

Um, no way

Even the 2nd one:

y=x+10
3=1+10
3=11

umm, no!

The reason its wrong is because sqrt(x^2+y^2=16) does not equal x+y=4... that is JUST WRONG. it cannot be done that way. No offense :)
x^2+y^2=16 is equal to x+y=4 (Spaure root the hole thing) plug in the second equation into thisone x+(x+10)=16 x+x+10=16 2x=6 6/2=3=x x=3 now plug the three into one ther equation 3+y=4 (subtract 3 from both sides) y=1 the answer is (3,1)

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