If you have a line passing through the middle of a circle, does it create a right angle at the intersection of the line and curve?

More generally, is it valid to define an angle created between a line and a curve? Is the tangent to the curve at the point of intersection a valid interpretation (I.e a semi circle has 2 right angles)

I saw it in a debate thread and it got me curious now.

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    $\begingroup$ As long as the derivative of both functions/curves is well-defined at that point, then yes. $\endgroup$ – Adam Rubinson Feb 27 at 8:57
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    $\begingroup$ Yes. "Angles determined by the intersection of two curves" (whether one of them is a line or not) are usually defined and computed in terms of angles between the respective tangent lines at the point of intersection. Complications ensue if there are no well-defined tangent lines, but frequently this is not the case. $\endgroup$ – leslie townes Feb 27 at 8:58

As you identify in your question, the real point of contention is the definition of angle. As the other answers have indicated, if your definition includes angles at the intersection of two curves then a semicircle certainly has two right angles.

However, as the controversy on Twitter (a few days ago at time of posting, started because this question was in someone's daughter's primary school homework) indicates, angle as commonly used is often restricted to the intersections of (straight) lines. Under this definition there is no angle at the vertices of a semicircle.

Which definition is more common is an empirical question (and the consensus seems to be that the wider definition is common among professional mathematicians, while the stricter definition might be something a layperson would describe), but to my mind the fact that there exist differing definitions makes your question ill-formed. I think the educational value is to expose the fact that we should be careful with how we think about our definition.

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    $\begingroup$ +1 for making the ambiguity a teaching moment. $\endgroup$ – Ethan Bolker Feb 28 at 0:19

Yes. The semi-circle can even be referred to sometimes as a ( Curvilinear) Diangle, sum of the two shown right angles is $\pi$.

If Q and P are opposite points of a diameter in a circle then the tangent at Q makes a right angle to the line PQ. Similarly tangent at P makes a right angle to the line QP.

The question is about two possibilities of the "diangle" and these two cases are possible, included in a rough sketch. The second region however cannot be be called a semicircle, but a lens shape etc..

enter image description here

  • $\begingroup$ In your third paragraph you seem to be understanding the OP's question as 'is a figure with two right angles necessarily a semicircle', which isn't the correct reading of the question (from the wording, as well as the school homework question on Twitter which the OP read about). $\endgroup$ – dbmag9 Feb 27 at 22:47
  • $\begingroup$ That was as much of the guessing I could do... $\endgroup$ – Narasimham Feb 27 at 22:54
  • $\begingroup$ Your question is certainly more interesting! But it's a different question. :) $\endgroup$ – dbmag9 Feb 27 at 23:07
  • $\begingroup$ To me it appears that is what OP asked. So to be sure I cut and pasted his comments below the two figures. Specifically the definitions included 1) tangent 2) intersection as represented.. $\endgroup$ – Narasimham Feb 27 at 23:33

More generally, is it valid to define an angle created between a line and a curve?

Yes. It can be done with a little calculus. Calculus gives us a workable definition of a unique tangent vector at each point on a curve. So, we can use calculus to translate questions about curves into equivalent questions about lines.

The line $\overline{AH}$ crosses the circle at $A$ and $H$. It is intuitively plausible that the line $\overline{GH}$ from the point $G$ on the circle to the point $H$ becomes perpendicular to $\overline{AH}$ as $G$ approaches $H$. If there is a line through the circle perpendicular to $\overline{AH}$ then it must be the line that $\overline{GH}$ approaches as $G$ goes to $H$. It is possible to rigorously define such a line using the mathematically precise definitions of limit and derivative from calculus.

The equation for the curve of the circle in polar coordinates is

$(x(\theta), y(\theta))=(r\cos(\theta), r\sin(\theta))$ with $r$ is constant.

The equation for the line through the circle in polar coordinates is

$(x(r), y(r))=(r\cos(\theta), r\sin(\theta))$ with $\theta$ is constant.

The vector tangent to the circle at $\theta$ is $\vec{a}=lim_{h\rightarrow 0} \frac{(x(\theta+h), y(\theta+h))-(x(\theta), y(\theta))}{h}$.

This is a mathematically rigorous definition of the notion of the tangent to the curve at a point.

$\vec{a}=\frac{d(x(r,\theta),y(r\theta))}{d\theta}=\frac{d(r\cos(\theta),(r\sin(\theta))}{d\theta}=(-r\sin\theta , r\cos\theta)$

Similarly, the vector tangent to the line at $r$ is

$\vec{b}=\frac{d(x(r,\theta),y(r\theta))}{dr}=\frac{d(r\cos(\theta),(r\sin(\theta))}{dr}=(\cos \theta , \sin\theta)$

The Dot Product.

$\vec{a} \cdot \vec{b}= |\vec{a}| |\vec{b}| \cos(\phi)$ where $\phi$ is the angle between $\vec{a}$ and $\vec{b}$.


$\vec{a} \cdot \vec{b}=0 \rightarrow \vec{a} \perp \vec{b}$.

$\vec{a} \cdot \vec{b} = -r \sin \theta \cos\theta + r \cos \theta \sin \theta=0$

Therefore, the vectors are orthogonal.

Assume the line is crossing the unit circle where $\theta=0$ then

$a=(0,1)|_{\text{where the line crosses}}$ and $b=(1,0)|_{\text{where the line crosses}}$.

And you can always define any circle to be a unit circle by changing your units of measurement. So this particular choice of $a$ and $b$ really holds for any circle.

  • $\begingroup$ This seems like overkill; if you're invoking calculus and trigonometric functions you could just as well invoke the fact that a radius of a circle is perpendicular to the tangent at that point; hence the tangents at the vertices of the semicircle are perpendicular to the straight edge. $\endgroup$ – dbmag9 Feb 27 at 23:24
  • $\begingroup$ @dbmag9 ok, but the notion of "tangent at a point" is rigorously defined through calculus. that's the derivative and requires the notion limit. So, this would have been a better answer if I mentioned that. I thought more people would answer. my answer wasn't intended to be the definitive answer. $\endgroup$ – Directions In Physics Feb 27 at 23:34
  • $\begingroup$ @dbmag9 thanks. i think i improved my answer based on your comment. now it's a real answer. still overkill though. $\endgroup$ – Directions In Physics Feb 27 at 23:49

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