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.

Argument | Reason | |
---|---|---|

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 |

- Q.E.D

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.

- Introduction to constructions
- Copy a line segment
- Sum of n line segments
- Difference of two line segments
- Perpendicular bisector of a line segment
- Perpendicular at a point on a line
- Perpendicular from a line through a point
- Perpendicular from endpoint of a ray
- Divide a segment into n equal parts
- Parallel line through a point (angle copy)
- Parallel line through a point (rhombus)
- Parallel line through a point (translation)

- Bisecting an angle
- Copy an angle
- Construct a 30° angle
- Construct a 45° angle
- Construct a 60° angle
- Construct a 90° angle (right angle)
- Sum of n angles
- Difference of two angles
- Supplementary angle
- Complementary angle
- Constructing 75° 105° 120° 135° 150° angles and more

- Copy a triangle
- Isosceles triangle, given base and side
- Isosceles triangle, given base and altitude
- Isosceles triangle, given leg and apex angle
- Equilateral triangle
- 30-60-90 triangle, given the hypotenuse
- Triangle, given 3 sides (sss)
- Triangle, given one side and adjacent angles (asa)
- Triangle, given two angles and non-included side (aas)
- Triangle, given two sides and included angle (sas)
- Triangle medians
- Triangle midsegment
- Triangle altitude
- Triangle altitude (outside case)

- Right Triangle, given one leg and hypotenuse (HL)
- Right Triangle, given both legs (LL)
- Right Triangle, given hypotenuse and one angle (HA)
- Right Triangle, given one leg and one angle (LA)

- Finding the center of a circle
- Circle given 3 points
- Tangent at a point on the circle
- Tangents through an external point
- Tangents to two circles (external)
- Tangents to two circles (internal)
- Incircle of a triangle
- Focus points of a given ellipse
- Circumcircle of a triangle

- Square given one side
- Square inscribed in a circle
- Hexagon given one side
- Hexagon inscribed in a given circle
- Pentagon inscribed in a given circle

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