# Construction help!

• Apr 6th 2013, 09:20 PM
Jammix
Construction help!
Given a line, L and two points, A & B not on the line, construct a circle which contains both points and is tangent to
the given line.
• Apr 8th 2013, 10:11 PM
ibdutt
Re: Construction help!
• Apr 9th 2013, 12:41 PM
MINOANMAN
Re: Construction help!
The lines AB and L intersect at a fixed point P.
The circle we search has centre at O . and the line L is tangent to this circle at the point Q.
According to a well known theorem of the Euclidean Geometry (PA) X (PB) = (PQ)^2 therefore PQ =SQRROOT[(PA) X (PB)]
this expression defines the position of the point of tangency Q along the line L . The centre of the circle is the intersection point O of the perpendicular bisector of AB and the perpendicular line to the line L at Q. OA =OB =OQ = r is the radius of the circle .

MINOAS
• Apr 9th 2013, 01:22 PM
Plato
Re: Construction help!
Quote:

Originally Posted by Jammix
Given a line, L and two points, A & B not on the line, construct a circle which contains both points and is tangent to the given line.

Quote:

Originally Posted by MINOANMAN
The lines AB and L intersect at a fixed point P.
The circle we search has centre at O . and the line L is tangent to this circle at the point Q.
According to a well known theorem of the Euclidean Geometry (PA) X (PB) = (PQ)^2 therefore PQ =SQRROOT[(PA) X (PB)]
this expression defines the position of the point of tangency Q along the line L . The centre of the circle is the intersection point O of the perpendicular bisector of AB and the perpendicular line to the line L at Q. OA =OB =OQ = r is the radius of the circle .

The original question contains a mistake.
What if your point is between $A~\&~B$, that is $A-P-B~?$ If so the no such circle exists.

So it must be stated that $A~\&~B$ are on the same side of $\ell$.

But then it may be that $\overleftrightarrow {AB} \cap \ell = \emptyset$, parallel. In which case $P$ does not exist.