This page shows how to draw the two possible tangents to a given circle through an external point with compass and straightedge or ruler. This construction assumes you are already familiar with Constructing the Perpendicular Bisector of a Line Segment.
This is the same drawing as the last step in the above animation with lines OJ and JM added.
|1||OM = MP = JM||M was constructed as the midpoint of OP (See Constructing the perpendicular bisector of a line segment for method and proof) and JM=OM because JM was constructed with compass width set from MO||2||JMO is an isosceles triangle||JM=OM from (1)||3||∠JMO = 180–2(∠OJM)||Interior angles of a triangle add to 180°. Base angles of isosceles triangles are equal.||4||JMP is an isosceles triangle||JM=MP from (1)||5||∠JMP = 180–2(∠MJP)||Interior angles of a triangle add to 180°. Base angles of isosceles triangles are equal.|
|6||∠JMP + ∠JMO = 180||∠JMP and ∠JMO form a linear pair|
|7||∠OJP is a right angle||
Substituting (3) and (5) into (6):
(180–2∠MJP) + (180–2∠OJM) = 180
Remove parentheses and subtract 360 from both sides:
–2∠MJP –2∠OJM = –180
Divide through by –2::
∠MJP + ∠OJM = 90
|8||JP is a tangent to circle O and passes through P||JP is a tangent to O because it touches the circle at J and is at right angles to a radius at
the contact point.
(see Tangent to a circle.)
|p||KP is a tangent to circle O and passes through P||As above but using point K instead of J|
The above animation is available as a printable step-by-step instruction sheet, which can be used for making handouts or when a computer is not available.