I need to find the angle at origin caused by two lines (one is the radius of one circle, and the other is tangent to the other circle).

Please see image below:

The Point A on the green circle is known, the dashed line and shorter full line begin at the origin. The other line is tangent to the red circle and passing through Point A.

How would I calculate the angle a?

I know this is probably a basic question but my head is burnt out at work and I just need reassurance before making an epic mistake.

  • 1
    $\begingroup$ The radius of the big circle is $\sqrt{2}$ but the radius $r$ of a small circle is unknown. The angle $\alpha$ depends on $r.$ $\endgroup$
    – kmitov
    Mar 13, 2014 at 10:24
  • $\begingroup$ The radius of the red circle...and also: are both circles centered at the origin? $\endgroup$
    – DonAntonio
    Mar 13, 2014 at 10:50

1 Answer 1


Since from the drawing it is clear that $\;r<\sqrt 2\;$, we have that if $\;\alpha\;$ is the wanted angle then

$$\cos\alpha=\frac r{\sqrt2}\implies \alpha=\arccos\frac r{\sqrt2}$$

  • $\begingroup$ to find the angle a do we simply do a = arccos(radius of small circle/radius of larger circle) $\endgroup$
    – Amie-lea
    Mar 13, 2014 at 11:10
  • $\begingroup$ @Amie-lea Well, yes...choosing accordingly the range of the angle. Isn't the above basic trigonometric formula clear enough? $\endgroup$
    – DonAntonio
    Mar 13, 2014 at 11:12
  • $\begingroup$ Yes I just wanted to clarify it as the point (1,1) was just added for simplicity. $\endgroup$
    – Amie-lea
    Mar 13, 2014 at 11:16
  • $\begingroup$ In fact the point $\;(1,1)\;$ has no importance whatsoever in this case, @Amie-lea : going by the same drawing, any other point on the outer circle gives the very same result. $\endgroup$
    – DonAntonio
    Mar 13, 2014 at 11:17
  • $\begingroup$ The same result with sqrt(2) on the bottom, or would this be replaced with the radius of the outer circle in other circumstances? Sorry for basic questions just want clarification $\endgroup$
    – Amie-lea
    Mar 13, 2014 at 11:25

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