# The equations {(y = c x^2+d, (c > 0, d < 0)),(x = a y^2+ b, (a > 0, b < 0)):} have four intersection points. Prove that those four points are contained in one same circle ?

Sep 14, 2016

See below.

#### Explanation:

Making the following transformations

$\left\{\begin{matrix}y = c {x}^{2} + d \\ x = a {y}^{2} + b\end{matrix}\right. \to \left\{\begin{matrix}\frac{y}{c} = {x}^{2} + \frac{d}{c} \\ \frac{x}{a} = {y}^{2} + \frac{b}{a}\end{matrix}\right. \to {x}^{2} + {y}^{2} - \frac{x}{a} - \frac{y}{c} + \frac{d}{c} + \frac{b}{a} = 0$

and comparing with the generic circle equation

$C \to {\left(x - {x}_{0}\right)}^{2} + {\left(y - {y}_{0}\right)}^{2} - {r}_{0}^{2} = 0$

$\left\{\begin{matrix}{x}^{2} + {y}^{2} - \frac{x}{a} - \frac{y}{c} + \frac{d}{c} + \frac{b}{a} = 0 \\ {x}^{2} + {y}^{2} - 2 {x}_{0} x - 2 {y}_{0} y + {x}_{0}^{2} + {y}_{0}^{2} - {r}_{0}^{2} = 0\end{matrix}\right.$

we conclude that with $\left(c > 0 , d < 0 , a > 0 , b < 0\right)$

$\left\{\begin{matrix}{x}_{0} = \frac{1}{2 a} \\ {y}_{0} = \frac{1}{2 c} \\ {r}_{0} = \sqrt{\frac{1}{2 a} ^ 2 + \frac{1}{2 c} ^ 2 - \left(\frac{d}{c} + \frac{b}{a}\right)}\end{matrix}\right.$

$C$ represents a circle which contains the solutions.

As an example, using the values

$a = 5 , b = - 2 , c = 3 , d = - 3$

we obtained the following outcome.