How to prove this locus? What is the locus of the midpoint of a line segment that is drawn from a given external point $P$ to a given circle with center $O$ and radius $r$ ?
Playing along with Geogebra I've noticed that the locus of these point is a circle with diameter $r$ but i dont know how to prove it .Can you please give me some hints (not solutions)?
(Only geometric methods,i.e no analitycal methods allowed)
Thanks.
EDIT:
@Jhon Douma 
 A: Let O be the centre of the circle and P the external point. Let points A and B be on the circle so that AB is a diameter and BOAP is a straight line. Let the midpoint of BP be B' and the midpoint of AP be A'.
Now consider any point C on the circle and let C' be the midpoint of CP. Then it it straightforward to show that triangle BCP is similar to triangle B'C'P and that triangle ACP is similar to triangle A'C'P. Both pairs are in ratio 2:1.
Now angle BCA is 90 therefore so is angle B'C'A'. Therefore C' lies on a circle whose diameter is A'B', with the required diameter equal to one half of AB 
A: Let the centre of the circle be $O$ and the external point be $P$. Let $Q$ be a point on the circle. Let the midpoint of $\overline{PQ}$ be $R$. 
Through $R$, draw a line parallel to $\overline{OQ}$ and let it meet $PO$ at $S$. What is the distance $\overline{RS}$?
A: HINT...start by parametrizing the point On the circle with $$(x,y)=(r\cos\theta,r\sin\theta)$$ then write down the midpoint in terms of $\theta$ which you can then eliminate...
A: You can assume the Cartesian co-ords for your point P (a, b) and centre of the circle O (h, k) . Then, you'll be able to derive an equation for the circle (easily). Then, take a certain point (x1,  y1) on the circle and it'll be really easy to derive the co-ords for the midpoint joining those two. Then apply the property that (x1,  y1) is on the circle. I think that'd be able to do it. 
