Question

Consider the circle (x-2)^2+(y+1)^2 = 16
--> would a solution to this equation be any point where the circle crosses the x axis? Is there algebra I should be using?

Answer 1

this is the equation of a circle with center (2,-1) and radius 4

Answer 2

yup that's as far as I got :P

Answer 3

because the statement
\[(x-2)^2+(y+1)^2=16\] means the distance between (x,y) and (2,-1) is 4

Answer 4

oh i see it crosses the x-axis where y = 0

Answer 5

a solution to this problem is anywhere that is on the circle (x-2)^2+(y+1)^2=16

basically you have a circle with radius 4 and center (2,-1)

Answer 6

so replace y by zero, solve for x

Answer 7

you get
\[(x-2)^2+1=16\]
\[(x-2)^2=15\]
\[x-2=\pm\sqrt{15}\]
\[x=2\pm\sqrt{15}\]

Answer 8

so the solution would be 2+1 root 15?

Answer 9

2+- I MEAN

Answer 10

so it crosses at two places,
\[(2+\sqrt{15},0)\] and
\[(2-\sqrt{15},0)\]

Answer 11

yes you have it!

Answer 12

cheers :)

Answer 13

You rock satellite!