SOLUTION: someone help me please! Thank you. Show all work too please. Find the point(s) of intersection of the line x - y = -6 and the circle x2 + y2 = 18 by solving the system of equa

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Question 972385: someone help me please! Thank you. Show all work too please.
Find the point(s) of intersection of the line x - y = -6 and the circle x2 + y2 = 18 by solving the system of equations.
Found 2 solutions by Alan3354, lwsshak3:
Answer by Alan3354(69443)   (Show Source): You can put this solution on YOUR website!
Find the point(s) of intersection of the line x - y = -6 and the circle x2 + y2 = 18 by solving the system of equations.
----------------
x - y = -6
y = x + 6
-----
x^2 + y^2 = 18
sub for y
x^2 + (x+6)^2 = 18
2x^2 + 12x + 36 = 18
x^2 + 6x + 9 = 0
Solved by pluggable solver: SOLVE quadratic equation (work shown, graph etc)
Quadratic equation (in our case ) has the following solutons:



For these solutions to exist, the discriminant should not be a negative number.

First, we need to compute the discriminant : .

Discriminant d=0 is zero! That means that there is only one solution: .
Expression can be factored:

Again, the answer is: -3, -3. Here's your graph:

================
There are always 2 solutions to a quadratic.
For this one, they are x = -3 and x = -3.
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Answer by lwsshak3(11628)   (Show Source): You can put this solution on YOUR website!
Find the point(s) of intersection of the line x - y = -6 and the circle x2 + y2 = 18 by solving the system of equations.
***
solve by substitution:
x-y=6
y=x-6
..
x^2 + y^2 = 18
x^2+(x-6)^2=18
x^2+x^2-12x+36=18
2x^2-12x+18=0
divide eq. by 2
x^2-6x+9=0 (perfect square)
(x-3)^2=0
(x-3)(x-3)=0
x=3 (multiplicity 2)
..
y^2=18-x^2
y^2=18-9=9
y=±√9=±3
y=3 (reject, does not check in first equation)
or
y=-3
..
Point of intersection (3, -3)
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