how do you find the x-intercepts and y-intercepts for this equation: x^2 + y^2 -12x + 8y + 27=0?

Question

what is it?

Answer 1

a previous poster noted that it was a circle with center at -6,-4 and radius 5. If you plot this though, you should notice that the circle does not touch the y axis, therefore there is no y intercept(in the real plane).

Now when you set y=0, you should get x=-3 And x=-9. Which are the two points the circle passes the x axis.

y=0 ->
(x+6)^2 +((0)+4)^2=25
(x+6)^2 + 16 = 25
(x+6)^2 = 9

Here is where many people do a mistake

sqrt = Square root.
so, take square root of both sides and you get

(x+6)= +3 AND -3, since (-3)*(-3) = 9, and (+3) * (+3) = 9

so,

x= +3 - 6
or
x =-3 -6

x=-3
x=-9

if you set x=0 you get (imaginary answers), which depending on your level of math, can mean the answer doesnt exist (or to be more precise, it does not exist in the real plane).

If it so happens your learning imaginary numbers, than you would need this answer which would be.

x=0 ->

y= +(sqrt(-11))
and
y= -(sqrt(-11))

you get this by doing the exact same steps as the y=0 one.

Answer 2

x^2+12x+36+y^2+8y+16-25=0
(x+6)^2+(y+4)^2=5^2
so it is a circle with centre -6,-4 and radius 5 units
so the intercepts are x=-9,+1
y intercepts=-11 and -1

Answer 3

x^2 + y^2 -12x + 8y + 27=0?

x-intercept (set x=0):
x^2 + y^2 -12x + 8y + 27=0?
y^2 + 8y + 27=0?

use the quadratic equation:
(-b +/- (b^2-4ac)^1/2)/2a
(-8 +/- (64 - 108)^1/2)/2
-4 +/- (44)^1/2 / 2
-4 +/- 3.317 = -7.317, -0.683

y-intercept (set y=0):
x^2 + y^2 -12x + 8y + 27=0?
x^2 - 12x + 27=0?

use the quadratic equation:
(-b +/- (b^2-4ac)^1/2)/2a
(12 +/- (144 - 108)^1/2)/2
6 +/- (36)^1/2 / 2
6 +/- 3 = 3, 9

x-intercepts: (0, -7.317), (0,-0.683)
y-intercepts: (3, 0), (9, 0)

Answer 4

Set x = 0 and solve for y.
Set y = 0 and solve for x.