1. going in circles

Man, I'm having trouble with existence proofs.

How do I prove...

If $|a|<2r$, then there exists two points of intersection for the two circles whose equations are $x^2+y^2=r^2$ and $(x-a)^2+y^2=r^2$?

I'm not sure how to express the two solutions backward. Also, the inequality is giving me grief. I want to know how to construct this proof from scratch. Not just use the given antecedent to justify the consequent.

I start by solving the system. But I have trouble. ...
$x^2+y^2=r^2\implies(x-a)^2=0$ Or $(\sqrt{r^2-x^2}-a)^2+\sqrt{r^2-x^2}^2=r^2\implies{}...$

2. Re: going in circles

From $x^2+y^2=r^2$ and $(x-a)^2+y^2=r^2$, it must be true that $x^2=(x-a)^2$ of $|x|=|x-a|$. This last inequality says the distance from x to a is the same as the distance from x to 0. Without a formal derivation, it is then clear that $x=a/2$. So with this x value,
$$y^2=r^2-a^2/4\text{ and so }4r^2-a^2\geq0\text{ or }2r\geq|a|$$
This last inequality is almost the assumption. You see that if $2r=|a|$, there is only one point of intersection. So with the assumption the two intersection points do exist and in fact are
$$(a/2,\sqrt{4r^2-a^2}/2)\text{ and }(a/2,-\sqrt{4r^2-a^2}/2)$$
Direct verification that $(a/2,\sqrt{4r^2-a^2}/2)$ is on both circles:
$$(a/2)^2+(4r^2-a^2)/4=a^2/4+r^2-a^2/4=r^2$$
and
$$(a-a/2)^2+(4r^2-a^2)/4=a^2/4+r^2-a^2/4=r^2$$
Similarly, the 2nd point is also on the two circles.

3. Re: going in circles

Upon reflection, I feel I must add to my previous post. First, for the conclusion to be true, $a\neq 0$. You said two circles, but this does not preclude the possibility that the two circles are the same.
Next, given two different circles $\Sigma_1$ and $\Sigma_2$ with respective centers and radii $C_1,r_1$ and $C_2,r_2$, let $d$ be the distance between the two centers.
1. One of the circles is contained within the other iff $d<|r_1-r_2|$.
2. If one circle is not contained within the other, the two circles intersect in at least one point iff $d\leq r_1+r_2$.
The above can be proved formally, but it should be fairly obvious by drawing pictures.

Existence proofs in general are not easy. The usual technique is to name specifically the supposed object, which is what I did in my original post. Frequently, you must name the sought object in terms of some other object known to exist. Example, if p is an odd prime, it is "known" that a primitive root b modulo p exists. Then a primitive root modulo $p^n$ exists by naming one to be $b+p$.

4. Re: going in circles

I think johng's answer, while correct, gets rather bogged down.

My approach would be to note that $x^2+y^2 = r^2 = (x-a)^2 + y^2$ so then, as johng noted $x^2=(x-a)^2$. This is easy to solve algebraically for $x=\tfrac{a}{2}$. Note that this confirms that the $x$ coordinate of every point of intersection is $x=\tfrac{a}{2}$.

Putting this into either circle equation yields $y^2+\left( \tfrac{a^2}{4}-r^2 \right)$. The discriminant of this quadratic in $y$ tells us that we have two real solutions iff $4r^2 - a^2 > 0$ or, equivalently, $(2r-a)(2r+a) > 0$. This gives $2r > a$ for $a \ge 0$ and $2r > -a$ for $a \le 0$. These combine to form $-2r < a < 2r$ which is the same as $|a| < 2r$. Note that $r$ is, by convention, positive.

5. Re: going in circles

fThat helped alot guys. Thanks. I'm working really hard right now on perfecting my understanding of proofs of mathematical statements and I'm tripping up on existence.

Basically this one goes...

1. Determine points of intersection of a circle by the backward method.
2. Determine the interval for which there are two solutions.
3. The proof: pick an arbitrary element within this interval and show that there are in fact two points of intersection.

Would you say that this is correct

Also, can we talk about the meaning of $|x|=|x-a|$? What can be gained from this? How is it that we know that this makes any sense at all?

6. Re: going in circles

I think that 3) is not necessary. My approach determined the coordinates of all intersections and then found the condition under which they are real and distinct.

7. Re: going in circles

Originally Posted by VonNemo19
If $|a|<2r$, then there exists two points of intersection for the two circles whose equations are $x^2+y^2=r^2$ and $(x-a)^2+y^2=r^2$?
The centres of the circles are $C_1 (0,0)~\&~C_2 (a,0)$ the midpoint of $\overline{C_1C_2}$ is $M:\left(\frac{a}{2},0\right)$
Because the points of intersection are on the perpendicular bisector of $\overline{C_1C_2}$.
\begin{align*}\left(\frac{a}{2}\right)^2+y^2&=r^2 \\y&=\pm\sqrt{r^2-\left(\frac{a^2}{4}\right)}\\\text{Therefore the points are}&:\left(\frac{a}{2},\pm\sqrt{r^2-\left(\frac{a^2}{4}\right)}\right) \end{align*}______note that $|a|\le 2r$ means ${\left(\frac{a^2}{4}\right)}\le r^2$, so the square root is real.