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 ?

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Answer 1 · 2016-09-14T14:35:18

See below.

Making the following transformations

${(y=cx^2+d),(x=ay^2+b):} ->{(y/c = x^2+d/c),(x/a=y^2+b/a):}->x^2+y^2-x/a-y/c+d/c+b/a=0$

and comparing with the generic circle equation

$C->(x-x_0)^2+(y-y_0)^2-r_0^2=0$

${(x^2+y^2-x/a-y/c+d/c+b/a=0),(x^2+y^2-2x_0x-2y_0y+x_0^2+y_0^2-r_0^2=0):}$

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

${(x_0=1/(2a)),(y_0=1/(2c)),(r_0=sqrt(1/(2a)^2+1/(2c)^2-(d/c+b/a))):}$

$C$ 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.

enter image source here

The equations {(y = c x^2+d, (c > 0, d < 0)),(x = a y^2+ b, (a > 0, b < 0)):}{(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 ?