# Systems of equations for circles

I was given a problem with a system of equations for two circles:$$\begin{cases}x^2+y^2=4\\(x-5)^2+y^2=4\end{cases}$$It’s clear that the two equations have no solution where both are true, but when I substitute the $$4$$ for $$x^2 + y^2$$ I get$$(x-5)^2 + y^2 = x^2 + y^2.$$ If I subtract the $$y^2$$ from each side, expand the $$(x-5)^2$$ and then subtract $$x^2$$ from each side I am left with$$-10x + 25 = 0,$$which can be solved as $$x = 2.5$$. This obviously doesn’t work if you enter it into the previous equations, so why did I get this solution?

• In solving $(x-5)^2+y^2=x^2+y^2$, you're answering the question "What points are equidistant from the centers of the circles?" (You've determined that all such points happen to lie on the line $x=2.5$, which is the perpendicular bisector of the centers.) This is different than answering the narrower question "What points are exactly $2$ units from each of the centers of the circles?" – Blue Jun 27 '20 at 2:00

Remember that an implication is true if the antecedent is false. You are given some equations as axioms. If the equations are inconsistent, eventually you can prove anything. That is why you check solutions in the original equations, because it is easy to assume something or make a transformation that is not a logical equivalence. Here when you subtracted $$y^2$$ from each side you lost the fact that it had to be greater than or equal to zero. Your solution is fine if $$x,y$$ can be complex. You find $$y=\pm 1.25i$$ and all is well. In the reals, not so much.
If $$x=2.5$$, and $$x^2+y^2=4$$, what is $$y$$?