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Using a ruler and a compass how can construct a line through a point and tangent to a circle. What I don't want is to eyeball the line by trying to line-up the ruler over the circle. Best if I could construct the point of intersection first and then draw the line.

PS. I know how to do it mathematically, I just don't know the steps for geometry, given A, C and the circle to find D.

Example

Update Based on answers here is the constructors. Thanks for the quick responses.

Solution

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Draw a Thales circle over the segment $AC$, it will intersect the desired $D$, because $AD\perp DC$:

  1. Draw the segment $AC$.
  2. Construct its midpoint $F$.
  3. Draw a circle with origin $F$ and radio $FA(=FC)$.
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You can do it with a straightedge alone, though the lines tend to clutter the scene.

Straightedge-only construction of polar and tangents

Start with the two solid black lines, with directions at your free disposal as long as you get four intersection points with the circle. (It helps to keep one line closer to the center of the circle and the other farther away.)

Then draw the dashed lines, then the blue line $p$. That blue line is called the polar of the point $P$. Interestingly, it does not depend on the particulars of the black lines you have begun with.

Now, if $P$ is outside the circle, then its polar $p$ crosses the circle, and the points of intersection are the points of tangency for tangents through $P$.

Bonus: This approach even works for a conic instead of a circle, as long as you are given at least five points of that conic. Takes even more lines though, unless the conic is drawn already, in which case it works the same way as for the circle. I have hinted at that elsewhere.

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  • $\begingroup$ Hey, I remember this from projective geometry. Take advantage of the pole-polar relashinship to find the line (in blue) connecting the tangent points. $\endgroup$ – ja72 May 29 '17 at 0:26
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Intersect the circle having $AC$ as a diameter with the initial circle: you will find the two points $D,D'$ such that $CD$ and $CD'$ are tangent to the initial circle. This comes from the fact that the circle is the locus of points that "see" any diameter under an angle equal to $\frac{\pi}{2}$.

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  1. From AC, find its midpoint F.
  2. Draw the circle using F as center and FA (or FC) as radius.
  3. The point(s) of intersection of the circles is D.
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  • $\begingroup$ Thanks, but how is this answer different from the one Berci provided? $\endgroup$ – ja72 Aug 12 '13 at 17:30
  • $\begingroup$ @ja72, Yes but have the extra description cut. $\endgroup$ – Mick Aug 22 '13 at 12:56

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