How do you find the x intercepts and y intercepts of a circle with a given circle besides graphing?

Question

The circle is not in the formula already.

The equation is x^2+y^2+4x-6y+4=0

Also I am not sure whether I should complete the square or not.

Answer 1

x-intercepts: (a,0)
a^2+0^2+4a-6(0)+4=0
a^2+4a+4=0
(a+2)^2 = 0
a = – 2

y-intercepts: (0,b)
0^2+b^2+4(0)-6b+4=0
b^2-6b+4=0
b = (6 ± √(36–16))/2
b = 3 ± √(9–4)
b = 3 ± √(5) ≈ 0.76393202250021, 5.23606797749979

Answer 2

Grace, for conic sections, the best thing to do is put the equations into standard form.  For a circle standard form is  (x - h)^2 + (y - k)^2 = r^2

 

where (h,k) is the center of the circle and r is the radius of the circle.

 

So, yes, you want to complete the square for x and y separately, and then pull all the remaining constant terms on the r-side of the equation to find the radius.  Note that the center is at (h,k) where the equation is x-h, and y-k, respectively; so you if you have a positive squared term, e.g. x+7, the center's x coordinate is -7.  I've provided the completing the square below, but it is fairly straightforward and I would encourage you to do it, and the final answer is provided as well.

 

Completing the square

x^2 + 4x + _ + y^2 - 6y + _ + 4 = 0

x^2 + 4x + 4 + y^2 - 6y + 9 + 4 -4 - 9 = 0

(x+2)^2 + (y-3)^2 - 9 = 0

(x+2)^2 + (y-3)^2 = 9

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

 

So Center is at (-2,3).

Now you want to solve this equation for y=0 (x-intercept) and for x=0 (y-intercept). [Thanks Steve].  At y=0, 

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

(x+2)^2=0

x=-2

 

At x=0,

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

 (y-3)^2 = 9-4

y=3±√5