Honestly, I am not even sure what formation the question is referring to. I am guessing that if the circles meet only at one point, then that point is a point that lies on the straight line that is formed between the centers of two circles. Is my guess correct?
$\begingroup$
$\endgroup$
1
-
$\begingroup$ I suggest you add "two different circles" instead of "two circles" because the two circles may overlap in which case both have many points in common (and all line passing by the center is your line of centers) $\endgroup$– PiquitoAug 24, 2016 at 0:33
Add a comment
|
3 Answers
$\begingroup$
$\endgroup$
The "line of centres" is simply the line passing through both circles' centres. The problem is relating to the positions of the circle intersections relative to this line, which we shall call $L$.
To prove the assertion, we note that two non-concentric circles can intersect in 0, 1 or 2 points. If the number of intersections is 0 or 1, we are done because neither side of $L$ can have more than one intersection. If it is 2, for each intersection there must be a corresponding one on the other side of $L$, simply because the entire figure is symmetric about $L$. This symmetry condition is only satisfied when the two intersections are on opposite sides of $L$, which implies at most one intersection on each side as asserted. Hence the statement in the title is true.
answered Aug 23, 2016 at 23:14
$\begingroup$
$\endgroup$
Let $A$ be the center of one circle having radius $r_1$, $B$ be the center of the second circle having radius $r_2$. Let $l$ be the line joining $A$ and $B$, and $C$ and $D$ be two points where the circles intersect lying on the same side of $l$. By construction $AC$ and $BC$ intersect at $C$, and $AD$ and $BD$ intersect at $D$.
Then $|AC|=|AD|=r_1$, $|BC|=|BD|=r_2$, and $|AB|=|AB|$ reflexively. Then by SSS ttiangkes $ABC$ and $ABD$ are congruent, and so are their corresponding angles.
Then angle $BAC$ is congruent with angle $BAD$ and so, with $C$ and $D$ on the same side of $l$ this implies line $AC$ and $AD$ coincide. Similarly angle $ABC$ is congruent eith angle $ABD$, and so line $BC$ coincides with line $BD$. Then the proposed four lines $AC, AD, BC, BD$ are only two distinct lines and can intersect at only one distinct point. The proposed two points $C$ and $D$ are both the same point.
answered Aug 24, 2016 at 0:32
$\begingroup$
$\endgroup$
HINT.- Taking the center line as $y$-axis in rectangular coordinates, you can consider the two circles of equations $x^2+y^2=r^2$ and $x^2+(y-k)^2=R^2$ taking with $0\lt k\lt r+R$ to ensure that there are intersection points. The solution of the system of the two equations gives two points in opposite sides of the $y$-axis.
