What is the x-intercept and y-intercept of the circle with the equation (x + 1)^2 + (y - 2)^2 = 16?

Answer 1

For the x-intercept you need to set y=0 which implies that (x+1)^2=12 and so x+1=sqrt(12) or x+1=-sqrt(12). Hence, there are two x-intercepts at the points (sqrt(12)-1,0) and (-sqrt(12)-1,0).

For the y-intercept you need to set x=0 which implies that (y-2)^2=15 and so y-2=sqrt(15) or y-2=-sqrt(15). Thus, there are two y-intercepts at the points (0,sqrt(15)+2) and (0,-sqrt(15)+2).

Answer 2

x-intercepts: set y = 0 and get (x + 1)² + 4 = 16

(x + 1)² = 12

x + 1 = ±2√3 So, x = -1 ± 2√3. The x-intercepts are (-1 + 2√3, 0) and (-1 - 2√3, 0).

y-intercepts: set x = 0 and get (y - 2)² = 15

y - 2 = ±√15 So, y = 2 ± √15. The y-intercepts are (0, 2 + √15) and (0, 2 - √15).