Question

Help please! Solve each system. In your work, identify the shape of the graph of each equation.
x^2+y^2+2x+2y=0
x^2+y^2+4x+6y+12=0

Answer 1

What does solve mean here?
Put the equation in standard conics form?

Answer 2

Does the equation have both x^2 and y^2 terms?
If yes, then it is NOT a parabola.
Here both x^2 and y^2 are in the equation. Therefore, it is not a parabola.

Do the coefficients of x^2 and y^2 terms have the same sign?
If yes then it is either an ellipse or a circle.
It is a circle if the coefficients of x^2 and y^2 are the same
(same sign, same numbers)
It is an ellipse if the coefficients of x^2 and y^2 are NOT the same
(same sign, but different numbers

If the coefficients of x^2 and y^2 terms have opposite signs then it is a hyperbola.

Here, in both cases, the coefficients of x^2 and y^2 are the same, each equal to +1 (same sign, same value). Therefore, both equations are circles.

Answer 3

If you need to find the center and radius of the circle then you need to put the equations in standard conics form:

(x-h)^2 + (y - k)^2 = r^2 where (h,k) is the center and r is the radius.

You put the equations in standard form by completing the square.

Answer 4

I know the answer for this problem, but I'm not sure how to show my work

Answer 5

The answers are (-2,-2) and (-6/5, -12/5)

Answer 6

x^2 + y^2 + 2x + 2y = 0
group x terms and y terms together
(x^ + 2x) + (y^2 + 2y) = 0
complete the square of (x^ + 2x) and (y^2 + 2y) separately
To complete the square of x^ + 2x, divide the x coefficient by 2: 2/2 = 1. This 1 will go inside the parenthesis to be square and you have to subtract the square of that term (-1^2):
x^ + 2x = (x + 1)^2 - 1^2 = (x + 1)^2 - 1

So the same with y^2 + 2y.
y^2 + 2y = (y + 1)^2 - 1

Put them together:
(x^ + 2x) + (y^2 + 2y) = 0 becomes
(x + 1)^2 - 1 + (y + 1)^2 - 1 = 0
(x + 1)^2 + (y + 1)^2 - 2 = 0
(x + 1)^2 + (y + 1)^2 = 2
(x + 1)^2 + (y + 1)^2 = (sqrt(2))^2

compare it to the standard conics form for a circle:
(x-h)^2 + (y - k)^2 = r^2 where (h,k) is the center and r is the radius.

So the center for (x + 1)^2 + (y + 1)^2 = (sqrt(2))^2 is:
(-1, -1) and the radius = sqrt(2).

Follow similar procedure for the second equation:
x^2+y^2+4x+6y+12=0
(x^2 + 4x) + (y^2 + 6y) + 12 = 0
complete the square of (x^2 + 4x) and (y^2 + 6y) separately:
x^2 + 4x. Take coefficient of x term and divide it by 2: 4/2 = 2. This 2 will go inside the parenthesis to be squared. You will have to subtract the square of 2.
x^2 + 4x = (x + 2)^2 - 2^2 = (x + 2)^2 - 4

Similarly, (y^2 + 6y) = (y + 3)^2 - 3^2 = (y + 3)^2 - 9

x^2+y^2+4x+6y+12=0 becomes:
(x + 2)^2 - 4 + (y + 3)^2 - 9 + 12 = 0
(x + 2)^2 + (y + 3)^2 - 1 = 0
(x + 2)^2 + (y + 3)^2 = 1
(x + 2)^2 + (y + 3)^2 = 1^2
compare it to the standard conics form for a circle: (x-h)^2 + (y - k)^2 = r^2 where (h,k) is the center and r is the radius.
center = (-2, -3) and radius = 1.

Both equations are circle.

Answer 7

I appreciate all of the work you did for this, but it is a system of conic systems type problem... I posted the correct answers, but i didn't see those answers in your work

Answer 8

x=-2 or -6/5 and y=-2 or -12/5

Answer 9

Did you learn what a row echelon form is (with matrices ?)

Answer 10

I have the answer, but I don't know how to show the answer

Answer 11

Oh, they are asking for the intersection points! Where do the two circles intersect?
They should have framed the question properly. What threw me off was "Solve EACH system." That means solve them separately. But what they actually want is for us to solve the two equations simultaneously.

x^2+y^2+2x+2y=0 ---- (1)
x^2+y^2+4x+6y+12=0 ---- (2)
subtract the equations:
-2x - 4y - 12 = 0
divide by -2
x + 2y + 6 = 0
x = -2y - 6 ------- (3)
Put this in (1):
(-2y-6)^2 + y^2 +2(-2y-6) + 2y = 0
4y^2 + 24y + 36 + y^2 -4y - 12 + 2y = 0
5y^2 + 22y + 24 = 0
5y^2 + 10y + 12y + 24 = 0
5y(y + 2) + 12(y + 2) = 0
(y + 2)(5y + 12) = 0
y = -2 or y = -12/5
Put y = -2 in (3)
x = (-2)(-2) - 6 = 4 - 6 = -2
So one point of intersection is: (-2, -2)

Put y = -12/5 in (3)
x = (-2)(-12/5) - 6 = 24/5 - 6 = (24 - 30) / 5 = -6/5
So the second point of intersection is: (-6/5, -12/5)
Or,
x = -2 or -6/5 and y = -2 or -12/5