Please help!!
x^2+y^2+6y+5=0
x^2+y^2-2x-8=0
Help with what?
You have given us two different circles.
What do you want to do?
They do intersect. Here is a picture of your problem
http://www.wolframalpha.com/input/?i=plot+x%5E2%2By%5E2%2B6y%2B5%3D0+,+x%5E2%2By%5E2-2x-8%3D0
you want
x^2+y^2+6y+5 = x^2+y^2-2x-8
6y+5 = -2x-8
y = -(2x+13)/6
Now use that to find x:
x^2 + (-(2x+13)/6)^2 - 2x - 8 = 0
40x^2 - 20x - 119 = 0
x = [5±9?15]/20
y = -3/20 (15±?15)
http://www.wolframalpha.com/input/?i=solve+x%5E2%2By%5E2%2B6y%2B5%3D0,+x%5E2%2By%5E2-2x-8%3D0
I'd be happy to help you! It seems like you have two equations involving variables x and y. These equations are quadratic equations in the form of x^2 + y^2 + bx + cy + d = 0, where b, c, and d are coefficients.
To find the solution to these equations, you can use a method called substitution or elimination. Let's solve them one by one.
Equation 1: x^2 + y^2 + 6y + 5 = 0
To solve this equation, we can complete the square to rewrite it in the form (x - h)^2 + (y - k)^2 = r^2, where (h, k) is the center of a circle and r is its radius.
First, let's isolate y:
x^2 + y^2 + 6y = -5
Now, complete the square for the y terms:
x^2 + (y^2 + 6y + 9) = -5 + 9
x^2 + (y + 3)^2 = 4
Now, we have the equation in the desired form. By comparing it to (x - h)^2 + (y - k)^2 = r^2, we can see that the center of the circle is (0, -3) and the radius is 2. Therefore, the solution to the first equation is a circle centered at (0, -3) with a radius of 2.
Equation 2: x^2 + y^2 - 2x - 8 = 0
Similarly, let's complete the square for this equation:
(x^2 - 2x) + y^2 = 8
Now, complete the square for the x terms:
(x^2 - 2x + 1) + y^2 = 8 + 1
(x - 1)^2 + y^2 = 9
Comparing it to (x - h)^2 + (y - k)^2 = r^2, we see that the center of the circle is (1, 0) and the radius is 3. Hence, the second equation represents a circle centered at (1, 0) with a radius of 3.
To summarize, you are dealing with two circles in the Cartesian coordinate system. The first circle is centered at (0, -3) with a radius of 2, and the second circle is centered at (1, 0) with a radius of 3.