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I don't understand why equating the equations of two circles doesn't give their intersection point. Instead, it seems to give the line passing through the intersection points, and you have to again use this equation within the equation of one of the circles in order to get the points through which the circle passes.

I find this rather strange, since in general, equating two functions gives their intersection points. Any thoughts on why this doesn't happen in the case of circles?

For example, you can try finding the intersection points of the following 2 circles: $$x^2+y^2=100$$and $$(x-6)^2+(y+5)^2=9$$

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    $\begingroup$ What does it mean to equate two equations? Normally you equation two expressions. $\endgroup$ – Potato Aug 10 '13 at 10:11
  • $\begingroup$ Ignoring some multi-value issues, you could write each circle as a function of $x$ and equate them. $\endgroup$ – Potato Aug 10 '13 at 10:13
  • $\begingroup$ Sorry, what I meant was, you substitute out whatever's possible using the 2nd equation. So, for example, you could substitute using $x^2+y^2=100$ in the remaining equation, and the $x^2$ and $y^2$ terms disappear. $\endgroup$ – Joebevo Aug 10 '13 at 10:13
  • $\begingroup$ Oh. Well this is certainly not a special case of equating two functions $f(x)$ and $g(x)$ as you seem to imply it is. It isn't unusual that you have to go back and forth between the equations to solve them. This even happens for linear equations. If you have $ax+by=c$ and $dx+ey=f$, you can solve for $x$ in one, substitute it in the other, and solve for the value of $y$. But you still have to go back and plug this value of $y$ in to find the value of $x$. A similar thing is going on here. A single substitution will give some, but not all, of the information you need. $\endgroup$ – Potato Aug 10 '13 at 10:19
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It sound like what you do is more or less to take the two equations $$ \tag{1} x^2+y^2 = 100 $$ $$ \tag{2} (x-6)^2 + (y-5)^2 = 9 $$ and subtracting them to get $$ \tag{3} (x-6)^2 - x^2 + (y-5)^2 - y^2 = 9 - 100 $$

Viewed in isolation, (3) is indeed the equation for a line through the two intersection points. However, it is to be expected that you loose information when you go from (1) and (2) to (3), because before you had two equations and now you have only one.

You also lose information when you equate two functions to learn where the two graphs intersect each other. In that case the information you lose is what the $y$ coordinate of the intersection point is; you need to go back to one of the original function definitions to find that once you have $x$.

In the "two functions" case, what the equation $f(x)=g(x)$ is a vertical line in the plane, since it doesn't mention $y$ at all. That's morally pretty close to the line (3) gives you -- the only difference is that the line from (3) may not be vertical.

Formally, we can say that what you do to get from (2) to (3) is to subtract something on both sides that you know to be the same, namely the two sides of (1). However, in order not to lose information, you have to remember the fact that what you subtracted was indeed equal, that is, your rewriting should leave you with (1) and (3) on the table, not with (3) alone.

Indeed, if you have (1) and (3) you can re-derive (2) by adding the two sides of (1) to (3) -- but (3) alone will not let you derive (1) nor (2).

What you do when you have (1) and (3) is that you can use (3) to substitute into (1) and find an equation for $x$ alone -- that is, a vertical line, and then you can substitute each solution for that back into (3) to find an equation in $y$ alone. Eventually you end up still having (two sets of) two equations, but now with the coordinates decoupled so you can read the solution off directly).

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