What are the x and y intercepts of x^2 + (y-4)^2 = 9 ?

x² + (y - 4)² = 9

When y = 0:
x² + (0 - 4)² = 9
x² = -7
x = ±√(-7)
The is no y-intercept for the curve.

When x = 0:
(y - 4)² = 9
y - 4 = 3 or y - 4 = -3
y = 7 or y = 1
y-intercepts are 1 and 7.

f(x,y) = x^2 + (y-4)^2 = 9...(1) defines a circle, with center at (0,4) and radius 3. This circle intersects the x axis when y = 0. For y = 0, (1)---> x^2 = 9-16 = -7...(2).
Clearly there exists no real x satisfying (2). Therefore there are no x-intercepts.
For x = 0, (1)---> (y-4)^2 = 0, ie., y = 4(+/-)3 = 1 or 7. There are therefore  2
y-intercepts. One at (0,1) and the other at (0,7).    

solving for x - intercept

set x = 0

x^2 + (y-4)^2 = 9
(0)^2 + (y - 4)^2 = 9
 y - 4 =± √(9)
y = 4  ± 3
 y = 4 + 3 = 7
 y = 4 - 3 = 1

y - intercept is (0,7), (0,1)..... Answer//

for x - intercept

set y = 0

x^2 + (0 - 4)^2 = 9
x^2 + 16 = 9
 x^2 = 9 - 16
x^2 = - 7
x = ± √(-7)
 x =± i√(7)

x- intercept ( i√(7), 0), (- i√(7), 0) Answer//

For x-intercepts put y=0, and for y intercepts x=0.
x intercepts: x^2+(0-4)^2=9
x^2+16=9
x^2=9-16= -7
Square of a real number can't be negative, which means there are no x-intercepts.
y intercepts.     0^2+(y-4)^2=9
y - 4 = +/- sqrt(9)
two solutions y1 and y2
y1 = 4 - 3 = 1,.    y2 = 4 + 3 = 7

With some experience you'll be able to notice that this equation can be written as
(x - 0)^2 + (y - 4)^2 = 3^2
Which is equation of a circle with center on (0,4) and radius 3.

hi yahoo has been hijacked by idiots