More precisely, do there exist two intersecting circles such that the tangent lines at the intersection points make an angle of 90 degrees, and one of the circles passes through the center of the other? I think the answer is no, and that this can only be the case if the circle passing through the center of the other circle is actually a line. I'm not sure how to show this though. Can someone give a proof or a counterexample?
Can two circles intersect each other at right angles, such that one circle passes through the center of the other circle
circlesconjecturesgeometry
Related Solutions
Let's say $\epsilon$ is the line defined by A and B and $K_1,K_2$ and $K_3$ the centers of the given circles.
Then we have the following 3 equations that will help us define $x_3,y_3,r_3$
$1$. $\sqrt{(x_3-x_2)^2+(y_3-y_2)^2}=r_3+r_2$ or $\sqrt{(x_3-x_2)^2+(y_3-y_2)^2}=|r_3-r_2|$
That is, the distance of the centers $K_2,K_3$ equals the sum of the radius $r_3,r_2$ if the circles are tangent outwardly or the $|r_3-r_2|$ if they're tangent inwardly.
$2$. $\sqrt{(x_3-x_1)^2+(y_3-y_1)^2}=r_3+r_1$ or $\sqrt{(x_3-x_1)^2+(y_3-y_1)^2}=|r_3-r_1|$
$3$. The distance of the $K_3$ from $\epsilon$ equals $r_3$ if you write $\epsilon :Ax+By+C=0$ then $r_3=\frac{|Ax_3+By_3+c|}{\sqrt{A^2+B^2}}$
Realize that $\angle NKM= 120^o$ and that $\angle KLN= 60^o$, which leaves $\Delta KLN$ an equilateral triangle. The answer is therefore $9\sqrt{3}$.
Related Question
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Best Answer
No. If their radii are $R$ and $r$, respectively, then the distance between the centres would be easy to calculate using the Pythagoras' theorem, and it will be:
$$\sqrt{R^2+r^2}$$
which is bigger than both $R$ and $r$, so neither of the circles can "reach" as far as the centre of the other.