Geometry (B): READ: Tangent Lines Reading

Tangent to a Circle

Tangent to a Circle Theorem:

A tangent line is always at right angles to the radius of the circle at the point of tangency.

Proof. We will prove this theorem by contradiction.

We start by making a drawing. $\overline{AB}$ is a radius of the circle. $A$ is the center of the circle and $B$ is the point of intersection between the radius and the tangent line.

Assume that the tangent line is not perpendicular to the radius.

There must be another point $C$ on the tangent line such that $\overline{AC}$ is perpendicular to the tangent line. Therefore, in the right triangle $ACB, \overline{AB}$ is the hypotenuse and $\overline{AC}$ is a leg of the triangle. However, this is not possible because $AC > AB$ . (Note that $AC =\;\mathrm{length\ of\ the\ radius} + DC$ ).

Since our assumption led us to a contradiction, this means that our assumption was incorrect. Therefore, the tangent line must be perpendicular to the radius of the circle. $\blacklozenge$

Since the tangent to a circle and the radius of the circle make a right angle with each other, we can often use the Pythagorean Theorem in order to find the length of missing line segments.

Example 1

In the figure, $\overline{CB}$ is tangent to the circle. Find $CD$ .

Since $\overline{CB}$ is tangent to the circle, then $\overline{CB}\perp\overline{AB}$ .

This means that $\triangle ABC$ is a right triangle and we can apply the Pythagorean Theorem to find the length of $\overline{AC}$ .

$(AC)^2 &= (AB)^2 + (BC)^2\\ (AC)^2 &= 25 + 64 = 89\\ AC &= \sqrt{89} \approx 9.43\ \text{in.}\\ CD &= AC - AD \approx 9.43 - 5 \approx 4.43\ \text{in.}$

Example 2

Mark is standing at the top of Mt. Whitney, which is $14,500\;\mathrm{feet}$ tall. The radius of the Earth is approximately $3,960\;\mathrm{miles.}$ (There are $5,280\;\mathrm{feet}$ in one mile.) How far can Mark see to the horizon?

We start by drawing the figure above.

The distance to the horizon is given by the line segment $CB$ .

Let us convert the height of the mountain from feet into miles.

$14500\ \text{feet} \times \frac {1\ \text{mile}}{5280\ \text{feet}} = 2.75\ \text{miles}$

Since $\overline{CB}$ is tangent to the Earth, $\triangle{ABC}$ is a right triangle and we can use the Pythagorean Theorem.

$(AC)^2 &= (CB)^2 + (AB)^2\\ (3960+2.75)^2 &= (CB)^2 + 3960^2\\ CB &= \sqrt{3962.75^2 - 3960^2} \approx 147.6\ \text{miles}$

Converse of a Tangent to a Circle

Converse of a Tangent to a Circle Theorem

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

Proof.

We will prove this theorem by contradiction. Since the line is perpendicular to the radius at its outer endpoint it must touch the circle at point $B$ . For this line to be tangent to the circle, it must only touch the circle at this point and no other.

Assume that the line also intersects the circle at point $C$ .

Since both $\overline{AB}$ and $\overline{AC}$ are radii of the circle, $\overline{AB} \cong \overline{AC}$ , and $\triangle ABC$ is isosceles.

This means that $m\angle{ABC} = m\angle{ACB} = 90^\circ$ .

It is impossible to have two right angles in the same triangle.

We arrived at a contradiction so our assumption must be incorrect. We conclude that line $BC$ is tangent to the circle at point $B$ . $\blacklozenge$

Example 3

Determine whether $\overline{LM}$ is tangent to the circle.

$\overline{LM}$ is tangent to the circle if $\overline{LM} \perp\overline{NM}$ .

To show that $\triangle{LMN}$ is a right triangle we use the Converse of the Pythagorean Theorem:

$(LM)^2 + (MN)^2 = 64 + 36 = 100 = 10^2$

The lengths of the sides of the triangle satisfy the Pythagorean Theorem, so $\overline{LM}$ is perpendicular to $\overline{MN}$ and is therefore tangent to the circle.

Tangent Segments from a Common External Point

Tangent Segments from a Common External Point Theorem

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

Proof.

The figure above shows a diagram of the situation.

Given: $\overline{AC}$ is a tangent to the circle and $\overline{BC}$ is a tangent to the circle
Prove: $\overline{AC} \cong \overline{BC}$

Statement	Reason
1. $\overline{AB}$ is tangent to the circle	1. Given
2. $\overline{AC} \perp\overline{OA}$	2. Tangent to a Circle Theorem
3. $\overline{BC}$ is a tangent to the circle	3. Given
4. $\overline{BC} \perp\overline{OB}$	4. Tangent to a Circle Theorem
5. $\overline{OA} \cong \overline{OB}$	5. Radii of same the circle
6. $\overline{OC} \cong \overline{OC}$	6. Same line
7. $\triangle AOC \cong \triangle BOC$	7. Hypotenuse-Leg congruence
8. $\overline{AC} \cong \overline{BC}$	8. Congruent Parts of Congruent Triangles are Congruent $\blacklozenge$

Example 4

Find the perimeter of the triangle.

All sides of the triangle are tangent to the circle.

The Tangent Segments from a Common External Point Theorem tells us that:

$& CE = FC = 7\ \text{cm} &&\\ & FA = AD = 8\ \text{cm} &&\\ & DB = BE = 12\ \text{cm} &&\\ & \text{The perimeter of the triangle} && = AF + FC + CE + EB + BD + AD \\ &&& = 8\ \text{cm} + 7\ \text{cm} + 7\ \text{cm} + 12\ \text{cm} + 12\ \text{cm} + 8\ \text{cm} = 54\ \text{cm}$

Example 5

An isosceles right triangle is circumscribed about a circle with diameter of $24\;\mathrm{inches}$ . Find the hypotenuse of the triangle.

Let’s start by making a sketch.

Since $\overline{EO}$ and $\overline{DO}$ are radii of the circle and $\overline{AC}$ and $\overline{AB}$ are tangents to the circle,

$\overline{EO} \perp\overline{AC}$ and $\overline{DO} \perp\overline{AB}$ .

Therefore, quadrilateral $ADOE$ is a square.

Therefore, $AE = AD = 12 \;\mathrm{in.}$

We can find the length of side $\overline{ED}$ by using the Pythagorean Theorem.

$(ED)^2 &= (AE)^2 + (AD)^2\\ (ED)^2 &= 144 + 144 =288\\ (ED) &= 12\sqrt{2}\ \text{in.}$

$\triangle{ADE}$ and $\triangle{ABC}$ are both isosceles right triangles, therefore $\triangle ADE$ $\sim$ $\triangle ABC$ and all the corresponding sides are proportional:

$\frac{AE} {AC} = \frac{AD} {AB} = \frac{ED} {CB}$

We can find the length of $\overline{EC}$ by using one of the ratios above:

$\frac{AE} {AC}= \frac{ED} {CB} \Rightarrow \frac{12} {12+x} = \frac{12 \sqrt{2}} {2x}$

Cross-multiply to obtain:

$24x &= 144\sqrt{2} + 12\sqrt{2}x\\ 24x - 12\sqrt{2}x &= 144\sqrt{2}\\ 12x(2 - \sqrt{2}) &= 144\sqrt{2}\\ x &= \frac{144\sqrt{2}}{12(2 - \sqrt{2})} = \frac{12 \sqrt{2}} {2 - \sqrt{2}}$

Rationalize the denominators:

$x &= \frac{12 \sqrt{2}} {2 - \sqrt{2}} \cdot \frac{2 + \sqrt{2}} {2 + \sqrt{2}} = \frac{24 \sqrt{2} + 24}{2}\\ x &= (12\sqrt{2} + 12) \;\text{ in.}$

The length of the hypotenuse is $\overline{BC} = {2}{x} = 24 \sqrt{2} + 24 \;\text{ in} \approx 58 \;\text{ in.}$

Corollary to Tangent Segments Theorem

A line segment from an external point to the center of a circle bisects the angle formed by the tangent segments starting at that same external point.

Proof.

Given:
- $\overline{AC}$ is a tangent to the circle
- $\overline{BC}$ is a tangent to the circle
- $O$ is the center of the circle

Prove
- $\angle{ACO} \cong \angle{BCO}$

Proof.

We will use a similar figure to the one we used we used to prove the tangent segments theorem (pictured above).

$\triangle AOC \cong \triangle BOC$ by $HL$ congruence.

Therefore, $\angle{ACO} \cong \angle{BCO}$ . $\blacklozenge$

Example 6

Show that the line $y=5-2x$ is tangent to the circle $x^2 + y^2 = 5$ . Find an equation for the line perpendicular to the tangent line at the point of tangency. Show that this line goes through the center of the circle.

To check that the line is tangent to the circle, substitute the equation of the line into the equation for the circle.

$x^2 + (5 - 2x)^2 &= 5\\ x^2 + 25 - 20x + 4x^2 &= 5\\ 5x^2 - 20x + 20 &= 0\\ x^2 - 4x + 4 &= 0\\ (x - 2)^2 &= 0$

This has a double root at $x = 2$ . This means that the line intersects the circle at only one point $(2, 1)$ .

A perpendicular line to the tangent line would have a slope that is the negative reciprocal of the tangent line or $m = \frac{1} {2}$ .

The equation of the line can be written: $y = \frac{1} {2} x + b$ .

We find the value of $b$ by plugging in the tangency point: $(2, 1).$

$1 = \frac {1} {2} (2) + b \Rightarrow b = 0$

The equation is $y = \frac{1} {2} x$ and we know that it passes through the origin since the $y-$ intercept is zero.

This means that the radius of the circle is perpendicular to the tangent to the circle.

Common External Tangents

Here is an example in which you might encounter the use of common external tangents.

Example 1

Find the distance between the centers of the circles in the figure.

Let’s label the diagram and draw a line segment that joins the centers of the two circles. Also draw the segment $\overline{AE}$ perpendicular the radius $\overline{BC}.$

Since $\overline{DC}$ is tangent to both circles, $\overline{DC}$ is perpendicular to both radii: $\overline{AD}$ and $\overline{BC}$ .

We can see that $AECD$ is a rectangle, therefore $EC = AD = 15\;\mathrm{in}$ .

This means that $BE = 25\;\mathrm{in} - 15\;\mathrm{in} = 10\;\mathrm{in}$ .

$\triangle ABE$ is a right triangle with $AE = 40\;\mathrm{in}$ and $BE = 10\;\mathrm{in}$ . We can apply the Pythagorean Theorem to find the missing side, $\overline{AB}$ .

$(AB)^2 &= (AE)^2 + (BE)^2\\ (AB)^2 &= 40^2 + 10^2\\ AB &= \sqrt{1700} \approx 41.2\ \text{in}$

The distance between the centers is approximately $41.2\;\mathrm{inches.}$

Lesson Summary

In this section we learned about tangents and their relationship to the circle. We found that a tangent line touches the circle at one point, which is the endpoint of a radius of the circle. The radius and tangent line are perpendicular to each other. We found out that if two segments are tangent to a circle, and share a common endpoint outside the circle; the segments are congruent.

The following questions are for your own review. The answers are listed below for you to check your work and understanding.

Review Questions

Determine whether each segment is tangent to the given circle:
Find the measure of angle .
Find the missing length:
Find the values of the missing variables
Find the perimeter of the pentagon:
Find the perimeter of the parallelogram:
Find the perimeter of the right triangle:
Find the perimeter of the polygon:
Draw the line $y = {3}{x} + 10$ and the circle $x^2 + y^2 = 10$ . Show that these graphs touch at only one point.