Points of intersection between parabola and circle?

Question

The exact question:

What is the greatest possible number of points of intersection between a parabola and a circle?

The answer is 4, but I don't know how to get it.

I tried equating the circle and parabola equations, but that gave me 2 points, not 4.

Thanks!

To Dr D, how did you get the parabola equation y = x^2 - 3? Isn't the equation in the form (y-k)^2=4a(x-h)? So if k and h are both 0, then the equation should be y^2=4ax...

Wow, the 2nd answer is very complicated, but thanks! This is actually an SAT II math question. I thought about it and figured out why 4 is the answer when I drew a circle on top of an ellipse and found that they can intersect 4 times.

Answer 1

The maximum is 4, but it's also possible to get 3,2 or 1. It depends on which equations you're working with.

If you worked with x^2 + y^2 = 1
And y = x^2,
then you'll only get two because these two curves only intersect twice.

Try x^2 + y^2 = 4
and y = x^2 - 3
That should give you 4.

*EDIT*
The parabola you gave is oriented with the x axis. It doesn't really matter.
Try x^2 + y^2 = 4
and x = y^2 - 3

Answer 2

"I tried equating the circle and parabola equations, but that gave me 2 points, not 4"

Oh, it'll give you four points. When you square root both sides, there's a plus AND a minus, both for the x's AND for the y's, leaving four solutions.

For an example graph, imagine a circle of radius 1, centered at the origin. x^2 + y^2 = 1

If you're using a graphing calculator, graph your circle with two functions, y = sqrt(1 - x^2) and y = -sqrt(1 - x^2)

Now make your parabola with a vertex one unit below the bottom of your circle, with (h, k) at (0, -2). We need to make the coefficient on our parabola large enuff (making the parabola skinny enuff) so that it hits our circle in four places.

How about y + 2 = 5(x - 0)^2, which is also written
y = 5x^2 - 2

I just picked 5 as a coefficient. But anything bigger than 2 will work.

Answer 3

You can work it out algebraically as a number of people above have shown, or you can simply draw a graph with a picture of a parabola and circle intersecting. It should be self evident that the most intersections you can get is 4.

Answer 4

Lets look at y = x² and x² + (y - 5)² = 16

Looking @ the circle equation:

x² + (x² - 5)² = 16
x² + x⁴ - 10x² + 25 = 16
x⁴ - 9x² - 9 = 0

let u = x², so
u² - 9u - 9 = 0

u = ½{9 ± √[81 - 4(1)(9)] }
u = 9/2 ± ½√[45]

Then we have u = x², so
x² = 9/2 + ½√[45] and x² = 9/2 - ½√[45]

so x = ±√{9/2 ± ½√[45]} and y = x²
x =
{
√{9/2 + ½√[45]}
√{9/2 - ½√[45]}
-√{9/2 + ½√[45]}
-√{9/2 - ½√[45]}
}
For a total of four solutions, hope this helps.

Answer 5

Plug one equation into the different. (one million) 2x + y = 7 y = 7 - 2x (7 - 2x)^2 - 6(7 - 2x) - 4x + 17 = 0 40 9 - 28x + 4x^2 - 40 two + 12x - 4x + 17 = 0 4x^2 - 20x + 24 = 0 4(x^2 - 5x + 6) = 0 (x - 3)(x - 2) = 0 x = 3 or x = 2 Plug back in: 2(3) + y = 7 y = one million 2(2) + y = 7 y = 3 (3, one million) and (2, 3) Similiar for different problems.