# Geometrical Drawing

Given are a circumference $(C, r)$, a point $P$, a line $S$ and a length $D$. Determine a line $T$ that passes through $P$ and has intersection with circumference at the points $A$ and $B$ so that the sum of the distances of $A$ and $B$ to the line $S$ is equal to $D$.

Could someone help me with this please?

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Can you fix the spellings and make it more mathematical? I mean, you're given a circle $C$ of radius $r$, a point $P$, a line $S$. Are these objects floating or bear some connection. Like, $P$ lies on the circumference of $C$ or $S$ is a chord of the circle or some such thing. In particular, a figure will be appreciated as well. –  user21436 Mar 10 '12 at 3:40
Also posted to MO, and closed there, but I think there was a useful comment posted there. –  Gerry Myerson Mar 10 '12 at 5:29
@GerryMyerson, could you share the link? –  draks ... Apr 5 '12 at 23:02
@draks, sorry, I can't find it there. I suspect it was deleted, by the system or by the poster. –  Gerry Myerson Apr 10 '12 at 12:54

Problem

Given point $P$, the circunference $\Gamma$ whose center is point $C$ and whose radius is $r$ and the length $D$, find out the line $t$, such that $P \in t$, $\Gamma \cap t = \{A, B\}$ and $d(A,s)+d(B,s)=D$. See fig.1.

Figure $1$

Solution (See fig. 2.)

Figure $2$

1. Draw the line $s'$ such that $s' \parallel s$, $d(s,s')= \frac{D}{2}$ and $s' \cap \Gamma \neq \emptyset$.

2. Draw the line segment $CP$ and mark its midpoint $M$ on it.

3. Draw the circunference $\Lambda$ whose center is $M$ and whose radius is $CM$.

4. Find out the point $N$, such that $\{N\}=s' \cap \Lambda$ and $N$ is inside the circle delimited by $\Gamma$.

5. Draw the line $t$, such that $N \in t$ and $P \in t$.

6. Find out the points $A$ and $B$, such that $\{A,B\}= t \cap \Gamma$.

Explanation

Note that $N$ is the midpoint of $AB$ and $\angle CNA = \angle CNP = \frac{\pi}{2}$.

As $N$ is midpoint of $AB$, we have:

$$d(A,s)+d(B,s)= 2d(N,s) \Rightarrow$$ $$d(A,s)+d(B,s)= 2 \left(\frac{D}{2}\right) \Rightarrow$$ $$d(A,s)+d(B,s)= D$$

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