Topics
11. Circles
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United States of America
Grades 9-12

11.02 Lines tangent to a circle

Lesson
Practice

Introduction

When working with circles, we encounter different types of special lines and segments. The first type of line we will explore is a tangent line. We will learn how to construct a tangent line to a circle and apply theorems related to tangents of a circle in this lesson.

Tangent lines

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.

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.

A pair of tangent lines that forms a circumscribed angle has the following property:

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.

Examples

Example 1

Draw a circle and construct a tangent to the circle.

Worked Solution
Create a strategy

A tangent to a circle intercepts a circle at one point, so let's choose any point, P, to be the point of tangency. If the circle has center, O, we can draw a ray through OP that extends outside the circle.

Then, we can use that ray and a compass to construct a line perpendicular to OP through point P. The constructed line will be through P and perpendicular to radius OP, which makes it a tangent by the converse of the tangent radius theorem.

Apply the idea

We can follow these steps to construct a tangent:

  1. Draw a circle and mark its center, O
  2. Draw a ray with endpoint at the circle center with length more than double the radius of the circle. Mark where it crosses the circle as P
  3. Set the compass to a width less than OP, then place the compass end at P and make two arcs - one inside the circle and one outside the circle, call these A and B
  4. Set the compass to a width greater than OP but less than AB and make an arc to one side of \overleftrightarrow{OP}
  5. Without changing the compass width, put the compass at point B and make an arc that intersects with the last arc
  6. Label this point C
  7. Using a straight edge, draw a line through C and P
  8. \overleftrightarrow{CP} is the tangent
Reflect and check

Notice in step 3 that we are creating point A and point B equidistant from point P. This makes P the midpoint of \overline{AB}. Then, steps 4 through 7 follow the procedure for creating a perpendicular bisector to \overline{AB}. Therefore, we know we have drawn a tangent line to circle O.

Example 2

Determine if \overline{MN} is tangent to circle O.

Worked Solution
Create a strategy

If \overline{MN} is tangent to circle O, then \overline{MN} should be perpendicular to \overline{ON} by the tangent radius theorem. If these lines are perpendicular, then \angle MNO would be a right angle, making the triangle a right triangle. We can use the converse of the Pythagorean theorem to determine if the triangle is a right triangle.

Apply the idea

The converse of the Pythagorean theorem says if {a^2+b^2=c^2}, then this is a right triangle. Let {a=44, b=24,} and c=50.

\displaystyle a^2\displaystyle =\displaystyle 44^2
\displaystyle =\displaystyle 1936
\displaystyle b^2\displaystyle =\displaystyle 24^2
\displaystyle =\displaystyle 576
\displaystyle c^2\displaystyle =\displaystyle 50^2
\displaystyle =\displaystyle 2500
\displaystyle a^2+b^2\displaystyle =\displaystyle 1936+576
\displaystyle =\displaystyle 2512

Therefore, a^2+b^2\neq c^2 which tells us the triangle is not a right triangle. This also shows \overline{MN} is not perpendicular to \overline{ON}, so \overline{MN} is not tangent to circle O.

Example 3

In the diagram below, point B is a point of tangency. Find the radius of circle A.

Worked Solution
Create a strategy

We are given \overline{BC} is tangent to circle A which means \angle ABC is a right angle. This means \triangle ABC is a right triangle, and we can use the Pythagorean theorem to solve for r.

Apply the idea
\displaystyle a^2+b^2\displaystyle =\displaystyle c^2State the Pythagorean theorem
\displaystyle r^2+24^2\displaystyle =\displaystyle \left(r+16\right)^2Substitute side lengths
\displaystyle r^2+576\displaystyle =\displaystyle r^2+32r+256Evaluate the exponents
\displaystyle 576\displaystyle =\displaystyle 32r+256Subtract r^2 from both sides
\displaystyle 320\displaystyle =\displaystyle 32rSubtract 256 from both sides
\displaystyle 10\displaystyle =\displaystyle rDivide 32 from both sides

This shows the radius of circle A is r=10.

Example 4

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}

Worked Solution
Create a strategy

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}

Apply the idea

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, so they form a circumscribed angle. 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.

Reflect and check

Note that since this is an isoceles right triangle, we could also have found the perimeter using trigonometric ratios.

Since it is isosceles, we know that BD = BA = 9 and m\angle A = m\angle D = 45\degree, so

\displaystyle \sin{45}\displaystyle =\displaystyle \dfrac{9}{AD}
\displaystyle AD\cdot \sin{45}\displaystyle =\displaystyle 9
\displaystyle AD\displaystyle =\displaystyle \dfrac{9}{\sin{45}}
\displaystyle \approx\displaystyle 12.73

So, the perimeter is 9 + 9 + 12.73 \approx 30.73 feet.

Example 5

In the figure shown, \overline{AC} and \overline{BC} are tangents to the circle, C is a circumscribed angle, and O is the center of the circle.

a

Solve for x. Justify your answer.

Worked Solution
Create a strategy

By the tangent radius theorem, we know that if a line (segment) is tangent to a circle, then it is perpendicular to the radius drawn to the point of tangency.

Apply the idea

We have been told that \overline{AC} is a tangent of the circle, so A is a point of tangency. This means that \overline{AC} \perp \overline{OA}, so \angle OAC is a right angle.

Similarly, we have that \overline{BC} is a tangent of the circle, so B is a point of tangency. This means that \overline{BC} \perp \overline{OB}, so \angle OBC is a right angle.

ACBO is a quadrilateral, so has an interior angle sum of 360 \degree. We know that the measures of the interior angles are 90 \degree, 90 \degree, 74 \degree, and x\degree.

x=360-90-90-74=106

b

Given AO=10 and BC=13.3, find the perimeter of ACBO. Justify your answer.

Worked Solution
Create a strategy

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)
Apply the idea

Segment \overline{AO} and segment \overline{BO} are congruent because both start in the center of the circle and end on the arc of the circle. They are both radii, so AO=BO=10.

Segments \overline{AC} and \overline{BC} are both tangents to the circle from a common point C, so they form a circumscribed angle. Therefore, by the congruent segment theorem, we know that \overline{AC} and \overline{BC} are congruent, so AC=BC=13.3.

\displaystyle P_{ABCO}\displaystyle =\displaystyle AB+BC+CO+OA
\displaystyle =\displaystyle 13.3+13.3+10+10Substitute known values
\displaystyle =\displaystyle 46.6Evaluate the addition
Idea summary

A line is tangent to a circle if and only if it is perpendicular to a radius of the circle. If the line is a tangent line, we can draw a third line to create a right triangle and use the Pythagorean theorem or trigonometric ratios to solve problems related to lengths and angles of the triangle.

Two tangent lines meet at a point exterior to the circle and form a circumscribed angle. Tangent segments that meet at a point exterior to the circle are congruent.

Outcomes

G.C.A.2

Identify and describe relationships among inscribed angles, radii, and chords. Include the relationship between central, inscribed, and circumscribed angles; inscribed angles on a diameter are right angles; the radius of a circle is perpendicular to the tangent where the radius intersects the circle.

G.C.A.4 (+)

Construct a tangent line from a point outside a given circle to the circle.

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