2
$\begingroup$

There is given a line (AB) , a point C in the line $ C \in (AB) $ and a circle $ \Omega $ Construct the circle $ \omega $ tangent with $ \Omega$ and also tangent in C with (AB) .It should be done with ruler and compass. I can draw (OC) that intercepts $ \Omega $ in E. Than I draw a perpendicular line with (OC) in E. I don't know what else to do.

enter image description here

$\endgroup$
  • $\begingroup$ It is not clear what you are trying to do here. Can you clarify your question? $\endgroup$ – N. Owad Nov 15 '16 at 17:14
  • $\begingroup$ I think it is impossible with a line and a circle put in that positions.. $\endgroup$ – MattG88 Nov 15 '16 at 17:19
  • $\begingroup$ I am trying to draw a circle $\omega$ tanget with the line (AB) in point C and also tangent with the circle $\Omega$ . Point C, line (AB) and $\Omega$ are alredy given $\endgroup$ – alana Nov 15 '16 at 17:22
  • $\begingroup$ You want to do that by just "compass and ruler" or by analytic geometry? $\endgroup$ – G Cab Nov 15 '16 at 17:28
  • $\begingroup$ It's possible to construct with ruler and compass if you're given a unit length. This construction only involves square-roots, addition, and division, all of which can be done via ruler and compass. $\endgroup$ – Hrhm Nov 15 '16 at 17:29
1
$\begingroup$

Imagining the circle we want expanding by the radius of $\Omega$, we see that a circle with the same center goes through the center of $\Omega$ and a $C$ shifted perpendicular to $AB$ by the radius of $\Omega$.

Draw line $CD$ perpendicular to $AB$ through $C$ and then draw the perpendicular bisector of the center of $\Omega$ and the shifted $C$. The circle we want is centered at the intersection of the two lines.

enter image description here

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ how can I prove this, please? $\endgroup$ – alana Nov 15 '16 at 17:53
  • $\begingroup$ sorry, I didn't see you had provided the same answer .. $\endgroup$ – G Cab Nov 15 '16 at 18:36

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.