# Thread: Point of intersection of circle and parabola

1. ## Point of intersection of circle and parabola

(x-2)^2+(y-3)^2=6 and y=-2x^2+14x-1 are the two equations, but I dont know how to start finding all the point of intersections. Trying substitution got me a quartic equation

x^2 - 4x + y^2 - 6y = -7 [1]

(-2x^2 + 14x - 1) = y [2]

x^2 - 4x + (-2x^2 + 14x - 1)^2 - 6(-2x^2 + 14x - 1) = -7
4x^4-56x^3+213x^2-116x+14=0

2. ## Re: Point of intersection of circle and parabola

Originally Posted by Ilikebugs
(x-2)^2+(y-3)^2=6 and y=-2x^2+14x-1 are the two equations, but I dont know how to start finding all the point of intersections. Trying substitution got me a quartic equation

x^2 - 4x + y^2 - 6y = -7 [1]

(-2x^2 + 14x - 1) = y [2]

x^2 - 4x + (-2x^2 + 14x - 1)^2 - 6(-2x^2 + 14x - 1) = -7
4x^4-56x^3+213x^2-116x+14=0
I don't think you will be able to solve that exactly without using special 4th degree methods. Solving numerically for x I get .1734347923 and .4501500255.

3. ## Re: Point of intersection of circle and parabola

Is there a way to solve it without going beyond quadratics?

4. ## Re: Point of intersection of circle and parabola

In general, you are going to get a quartic. In particular cases, you can get other things happening. For example, given $(x-2)^2+(y-3)^2=6$ and $y=-2x^2+8x-1$ you can get a quadratic in $y$ which then leads to two quadratics in $x$. In other cases you'll get a quartic, but you'll be able to factorise it via the factor theorem and polynomial division.

5. ## Re: Point of intersection of circle and parabola

Originally Posted by Ilikebugs
Is there a way to solve it without going beyond quadratics?
I doubt it. Where did you get this problem? If it was from a typical calculus text I would expect a misprint or typo. Such texts usually give problems like this that are designed to work out easily with nice solutions.