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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?

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    $\begingroup$ We can fix one of the circles, say as the unit circle about the origin, and also the ray connecting their centers, say, the positive $x$-axis. Then just use coordinate geometry to prove your conjecture (in case of lack of any better idea). $\endgroup$
    – Berci
    Commented Nov 10, 2021 at 13:54
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    $\begingroup$ The answer is "yes". See this topology lecture for a counterexample. $\endgroup$
    – K. A. Buhr
    Commented Nov 10, 2021 at 22:58

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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.

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No, this is not possible.
Here's a hint. Consider the following points:

  • $A$: center of first circle (say radius $r_1$),
  • $B$: center of second circle (say radius $r_2$),
  • $C$: one point of intersection of the two circles.

Then, by the right-angle condition, you see that $ABC$ is a right-angled triangle. However, note that the sides are $r_1$, $r_2$, and $r_2$ with the right-angle being in between the sides $r_1$ and $r_2$. Check that this is not possible.

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If circle $C_2$ passes through the center of circle $C_1$ then the line passing through the center of $C_1$ and the intersection point of the circles intersects $C_1$ at a right angle.

On the other hand, that line is a secant line for $C_2$ and therefore not tangential to $C_2$.

enter image description here

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Let \begin{align} \vec{r}_1(\alpha)&=R\left(\cos \alpha, \sin \alpha\right)\\ \vec{r}_2(\beta)&=\lambda R\left(1+\cos \beta, \sin \beta\right) \end{align} be equations of the given circles where $\lambda$ and $R$ are both positive. The parameters $\alpha$ and $\beta$ span from $0$ to $2\pi$.

Assume their intersections occur at $\alpha = a$ and $\beta = b$, we have \begin{cases} R\cos a = \lambda R \left(\cos b + 1\right)\\ R\sin a = \lambda R \sin b \end{cases}

With some relatively easy algebra manipulations, we have the solution for the upper intersection as follows. \begin{cases} \cos a = \frac{1}{2\lambda}\\ \cos b = \frac{1-2\lambda^2}{2\lambda^2}\\ \sin a=\frac{\sqrt{4\lambda^2 -1}}{2\lambda}\\ \sin b =\frac{\sqrt{4\lambda^2 -1}}{2\lambda^2} \end{cases}

Next, find the direction vector of the tangents as follows.

\begin{cases} \vec{t}_1=\left.\frac{\mathrm{d}\vec{r}_1}{\mathrm{d}\alpha}\right|_{\alpha=a}=R\left(-\sin a, \cos a\right)\\ \vec{t}_2=\left.\frac{\mathrm{d}\vec{r}_2}{\mathrm{d}\beta}\right|_{\beta=b}=\lambda R\left(-\sin b, \cos b\right) \end{cases}

Now find the angle between the tangents.

\begin{align} \vec{t}_1 \cdot \vec{t}_2 &= |\vec{t}_1| |\vec{t}_2| \cos \theta \\ \lambda R^2 \left(\cos a \cos b + \sin a \sin b \right) &= \lambda R^2 \cos \theta\\ \cos \theta &=\cos a \cos b + \sin a \sin b\\ &= \frac{1}{2\lambda} \times \frac{1-2\lambda^2}{2\lambda^2} + \frac{\sqrt{4\lambda^2 -1}}{2\lambda}\times \frac{\sqrt{4\lambda^2 -1}}{2\lambda^2}\\ &= \frac{1-2\lambda^2+4\lambda^2-1}{4\lambda^3}\\ &=\frac{2\lambda^2}{4\lambda^3}\\ &=\frac{1}{2\lambda} \end{align}

They can be perpendicular each other (or $\theta=90^\circ$) if one of the circles has an infinite radius ($\lambda \to \infty$).

enter image description here

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Two such circles:

$\Gamma1 \begin{cases}x^2 + y^2 = 4 \\ z = 0\end{cases}$

$\Gamma2 \begin{cases}(x-1)^2 + z^2 = 1 \\ y = 0\end{cases}$

$\Gamma2$ passes through the centre of $\Gamma1$, and they intersect at right angles. Note that I could make $\Gamma2$ arbitrarily larger, including the same size as $\Gamma1$, and still fulfill the conditions, by shifting the centre away from $z=0$.

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