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9.02 Lines tangent to a circle

Lesson

Concept summary

A line that touches a circle at exactly one point is called a tangent line. The point of intersection where a tangent line touches the circle is called the point of tangency. If the tangent line is in the same plane as the circle, it is known as a tangent of a circle.

Tangent radius theorem

If a line is tangent to a circle, then it is perpendicular to the radius drawn to the point of tangency.

Circle C with a line tangent to the circle drawn. A radius is drawn to the point of tangency. This radius is perpendicular to the tangent line.
Converse of the tangent radius theorem

If a line is perpendicular to a radius of a circle at its endpoint on the circle, then the line is tangent to the circle

A segment of a tangent with one endpoint on the circle is called a tangent segment.

Any point exterior to a circle is the intersection point of exactly two tangent lines to the circle. This pair of tangent lines have the following properties:

Circumscribed angle

The angle formed at the point of intersection of two tangent lines to a circle

Circle C with two tangent lines drawn. The tangent lines intersect at a point outside the circle, and angle formed by the lines is marked.
Congruent segment theorem

If two segments from the same exterior point are tangent to a circle, then they are congruent.

Circle C with two congruent segments tangent to the circle. The tangents intersect at a point outside the circle.

Worked examples

Example 1

In the figure shown, \angle RMG and \angle RLG are right angles, and G is the center of the circle.

Circle G with radii G M and G L and a point R outside the circle. A segment is drawn from M to R, and from L to R.
a

Identify all segments which are tangent to the circle.

Approach

By the converse of the tangent radius theorem, we know that if a line (or segment) is perpendicular to the radius of a circle, and meets the radius at its endpoint on the circle, then the line (or segment) is tangent to the circle.

Solution

We have been told that \angle RMG is a right angle, and we know that \overline{MG} is a radius of the circle since G is the center. So by the converse of the tangent radius theorem, \overline{RM} must be a tangent segment.

Similarly, we have that \angle RLG is a right angle, and \overline{LG} is a radius of the circle. So by the converse of the tangent radius theorem, \overline{RL} must also be a tangent segment.

b

Identify segment pairs which have the same length.

Approach

We need to use what we know about circles and tangent lines:

  • Any line from the center of the circle to the arc of the circle has length r radius.
  • Tangent segments from a common endpoint not on the circle are congruent (congruent segment theorem)

Solution

Segment \overline{GM} and segment \overline{GL} have the same length because both start in the center of the circle and end on the arc of the circle. They are both radii.

Segments \overline{MR} and \overline{LR} are both tangents to the circle from a common point R. Therefore, by the congruent segment theorem, we know that \overline{MR} and \overline{LR} have the same length.

Example 2

An isosceles right triangle \triangle{ABD} is constructed such that the side \overline{BD} passes through the center C of a circle and the hypotenuse \overline{AD} is tangent to the circle at E.

Circle C, right triangle A B D, and a point E on circle C. Angle A B D is a right angle. Side A B is tangent to circle C at B. Side A D is tangent to circle C at E. Side B D passes through C. A B has a length of 9 feet, and E D has a length of 3.7 feet.

If AB measures 9 feet and ED measures 3.7 feet, determine the perimeter of \triangle{ABD}

Approach

We know that C is the center of the circle, so \overline{BC} is a radius of the circle. Since \angle ABD is a right angle, the converse tangent radius theorem tells us that \overline{AB} must be tangent to the circle.

Since \triangle{ABD} is isosceles, we also have that \overline{AB} \cong \overline{BD}.

Finally, we know that \overline{AE} is tangent to the circle.

We can put all of this together with the congruent segment theorem to determine the perimeter of \triangle{ABD}

Solution

We are given that AB = 9 feet.

Since \triangle{ABD} is isosceles, with \overline{AB} \cong \overline{BD}, this means that BD = 9 feet as well.

We also know that \overline{AB} and \overline{AE} are both tangent to the circle, from a common exterior point A. By the congruent segment theorem, these segments must be equal in length, and therefore we also have that AE = 9 feet.

So we have:

\displaystyle \text{Perimeter of}\, \triangle{ABD}\displaystyle =\displaystyle AB + BD + AE + ED
\displaystyle =\displaystyle 9 + 9 + 9 + 3.7
\displaystyle =\displaystyle 30.7

So the perimeter of \triangle{ABD} is 30.7 feet.

Reflection

Note that since this is a right triangle, we could also have found the perimeter using the Pythagorean theorem.

Since it is isosceles, we know that BD = BA = 9, and so

\displaystyle AD^2\displaystyle =\displaystyle AB^2 + BD^2
\displaystyle =\displaystyle 9^2 + 9^2
\displaystyle =\displaystyle 2 \cdot 9^2
\displaystyle AD\displaystyle =\displaystyle \sqrt{2 \cdot 9^2}
\displaystyle =\displaystyle 9 \sqrt{2}
\displaystyle \approx\displaystyle 12.73

So the perimeter is 9 + 9 + 12.73 = 30.73 feet.

Outcomes

MA.912.GR.5.5

Given a point outside a circle, construct a line tangent to the circle that passes through the given point.

MA.912.GR.6.1

Solve mathematical and real-world problems involving the length of a secant, tangent, segment or chord in a given circle.

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