X^2+y^2+6y+3=0
x^2+y^2-9x-6=0
List the solutions:
Please and thank you!
You have two circles:
x^2 + (y+3)^2 = 6
(x - 9/2)^2 + y^2 = 105/4
Naturally, they intersect where
x^2+y^2+6y+3 = x^2+y^2-9x-6
6y = -9x-9
y = -(3x+3)/2
Subbing that into one of the equations,
x^2+(3x+3)^2/4-9x-6=0
13x^2 - 18x - 15 = 0
x = 1/13 (9±2√69)
y = -3/13 (11±√69)
Hmm. I had expected a bit simpler answer.
To find the solutions to these equations, we can use a method called the quadratic formula. The quadratic formula is used to solve quadratic equations of the form ax^2 + bx + c = 0.
Let's start with the first equation: x^2 + y^2 + 6y + 3 = 0.
1. Rearrange the equation to isolate the x terms and the y terms on separate sides:
x^2 + y^2 + 6y = -3
2. Complete the square for the y terms by adding (6/2)^2 = 9 to both sides of the equation:
x^2 + y^2 + 6y + 9 = -3 + 9
x^2 + (y + 3)^2 = 6
3. Rewrite the equation in the standard form:
(x - 0)^2 + (y + 3)^2 = (sqrt(6))^2
The equation represents a circle centered at the origin (0, -3) with a radius of sqrt(6).
Now, let's move on to the second equation: x^2 + y^2 - 9x - 6 = 0.
1. Rearrange the equation to isolate the x terms and the y terms on separate sides:
x^2 + y^2 - 9x = 6
2. Complete the square for the x terms by adding (9/2)^2 = 81/4 to both sides of the equation:
x^2 - 9x + 81/4 + y^2 = 6 + 81/4
(x - 9/2)^2 + y^2 = 105/4
3. Rewrite the equation in the standard form:
(x - 9/2)^2 + (y - 0)^2 = (sqrt(105)/2)^2
The equation represents a circle centered at (9/2, 0) with a radius of sqrt(105)/2.
To list the solutions, we can find the points where these circles intersect. However, as the equations are in the form of circles, they might not intersect at any point. To determine the points of intersection, we would need to solve the equations simultaneously, which is beyond the explanation of this program.