Intersection of two circles giving reversed answer

Question

A little help needed. I need to derive the formula for the intersection points of two random circles. Equation: $\ x^2 + y^2 = 2.4^2$ and $ x^2 + (y+4)^2 = 17.16 $ I derived the equation which matches the equation from this site : http://mathworld.wolfram.com/Circle-CircleIntersection.html

However, an odd thing is happening. The x and y coordinates are reversed, as graphically visible. According to link, $\ x=(42−17.16+2.42)/−8=−0.575$ and $\ y^2=[(4∗16∗2.42)−(16−17.16+2.42)]/(4∗16)=2.33^2 $ so $\ y = 2.33$

However solving it on Wolframalpha gives the solutions as (−2.33, -0.575) and (2.33, -0.575). Could someone please explain using this method. Can some please explain how to solve this and why it is happening? Is the formula given in the link correct for all circles or only certain cases?

Answer 1

$x^2 + y^2 = 2.4^2\\ x^2 = 2.4^2 - y^2\\ x^2 + (y-4)^2 = 17.16\\ 2.4^2 - y^2 + (y-4)^2 = 17.16\\ 2.4^2 +8y+16 = 17.16\\ 8y = -4.6\\ y = -0.575$

Somewhere in there you transpose your x and y. Where?

$x = \pm \sqrt{2.4^2-0.575^2}$

Answer 2

That should have been: $$\ x^2 + y^2 = 2.4^2 ;\, y^2 + (x+4)^2 = 17.16 $$ to get answer as $ x= \pm 2.33 $