Determine the point of intersection between a circle (x+2)^2 + (y-2)^2 =1 and the parabola y=-(x+2)^2 +3.can anyone please outline the steps to solve it I have tried solve by substitution but I get Stuck
just plug and chug.
(x+2)^2 + (y-2)^2 =1
so (x+2)^2 = 1-(y-2)^2
y = -(x+2)^2+3
y = (y-2)^1-1 + 3
y = y^2-4y+4+2
y^2-5y+6 = 0
(y-3)(y-2) = 0
So the curves intersect at (-2,3) and (-2±1,2)
http://www.wolframalpha.com/input/?i=plot+(x%2B2)%5E2+%2B+(y-2)%5E2+%3D1,+y%3D-(x%2B2)%5E2+%2B3
Thanks Steve you saved my day
To determine the point of intersection between the circle and the parabola, we need to solve their equations simultaneously. Here are the steps to solve it:
Step 1: Set up the equations
The equation of the circle is (x+2)^2 + (y-2)^2 = 1.
The equation of the parabola is y = -(x+2)^2 + 3.
Step 2: Substitute the equation of the parabola into the equation of the circle
Replace y in the equation of the circle with -(x+2)^2 + 3:
(x+2)^2 + (-(x+2)^2 + 3 - 2)^2 = 1
Step 3: Simplify the equation
Expand and simplify the equation obtained from the substitution:
(x+2)^2 + ((-x-1)^2)^2 = 1
(x+2)^2 + (x^2 + 2x + 1)^2 = 1
Step 4: Solve the equation
Expand and collect like terms in the equation:
(x^2 + 4x + 4) + (x^4 + 4x^3 + 4x^2 + 4x^2 + 16x + 16 + 6x + 9) = 1
x^4 + 4x^3 + 9x^2 + 24x + 29 = 1
Rearrange the equation:
x^4 + 4x^3 + 9x^2 + 24x + 28 = 0
Step 5: Solve the equation for x
Unfortunately, solving this quartic equation directly can be quite complex and may not yield an easily interpretable solution. However, you can use numerical methods or graphing techniques to find approximate solutions.
Step 6: Substitute values of x back into the equation of the parabola
For each value of x found in Step 5, substitute them back into the equation of the parabola: y = -(x+2)^2 + 3.
This will give you the corresponding y-values for the points of intersection.
By following these steps, you can find the approximate points of intersection between the given circle and parabola.