SOLUTION: for the circle (x-2)^2+(y+3)^2=9 find the x and y intercepts. i need step by step

Algebra ->  Quadratic-relations-and-conic-sections -> SOLUTION: for the circle (x-2)^2+(y+3)^2=9 find the x and y intercepts. i need step by step      Log On


   

Question 294735: for the circle (x-2)^2+(y+3)^2=9 find the x and y intercepts. i need step by step
Found 2 solutions by Fombitz, stanbon:
Answer by Fombitz(32388) About Me  (Show Source):
You can put this solution on YOUR website!
To find the x-intercepts, set y=0 and solve for x.
%28x-2%29%5E2%2B%28y%2B3%29%5E2=9
%28x-2%29%5E2%2B%280%2B3%29%5E2=9
%28x-2%29%5E2%2B9=9
%28x-2%29%5E2=0
x-2=0
x=2
(2,0)
To find the y-intercepts, set x=0 and solve for y.
%28x-2%29%5E2%2B%28y%2B3%29%5E2=9
%280-2%29%5E2%2B%28y%2B3%29%5E2=9
4%2B%28y%2B3%29%5E2=9
%28y%2B3%29%5E2=5
y%2B3=0+%2B-+sqrt%285%29
y=+-3+%2B-+sqrt%285%29+
(0,+-3-sqrt%285%29+) and (0,-3%2Bsqrt%285%29)
.
.
.

Answer by stanbon(75887) About Me  (Show Source):
You can put this solution on YOUR website!
for the circle (x-2)^2+(y+3)^2=9 find the x and y intercepts.
i need step by step
---------------------------
x-intercept = ?
Let y = 0 and solve for "x":
(x-2)^2 + 3^2 = 9
(x-2)^2 = 0
x = 2 with multiplicity two.
That means the circle touches the x-axis at x=2,
but does not pass through it.
------------------
y-intercept = ?
Let x = 0 and solve for "y":
(-2)^2 + (y+3)^2 = 9
(y+3)^2 = 5
y+3 = +-sqrt(5)
y = -3+sqrt(5); y = 3-sqrt(5)
===================================
Cheers,
Stan H.