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.

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  • $\begingroup$ 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

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.

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  • $\begingroup$ 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

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