Which point lies outside the circle with equation (x-2)^2 +(y+3)^2=4?

Answer 1

the standard equation for a circle is
(x-x0)^2+(y-yo)^2 = radius^2

just look at the 4. the square root of 4 is 2.

so it is a circle with radius of 2.

It is centered at the point x0, y0.
which in your case is x0=2
and y0=-3

you can just remember to reverse the signs of the 2 and 3, or you can set up the equation and solve, like below, I just remember to flip them

(x-2)^2=0 (y+3)^2=0
(x-2)=0 (y+3)=0
x=2 y=-3

so any point outside of the circle centered at x=2, y=-3, and radius=2

hope it hels

Answer 2

Any (x - 2)^2 + (y + 3)^2 > 4

Answer 3

an absolutely infinitely large number of points.