# Lesson Explainer: Tangents of a Circle Mathematics

In this explainer, we will learn how to use the properties of tangents of circles to find missing angles or side lengths.

Recall that a tangent to a circle is a straight line that passes through exactly one point of the circle. The line does not go into the circle, but it just passes through the perimeter of the circle as seen in the diagram below.

We begin with an important theorem about the angle created by a tangent and a radius of a circle.

### Theorem: Angle between a Tangent and a Radius of a Circle

Any tangent to a circle is perpendicular to the radius at the point of contact.

The proof of this theorem relies on the fact that the shortest distance between a line and a point is the perpendicular distance between the two objects. In other words, the shortest line segment from the point to the given line must intersect perpendicularly with the line.

If a line is tangent to a circle, any point of the line is outside the circle, except the point of contact that lies on the circle. We know that the distance between the center of a circle and a point exterior to the circle must be greater than the radius of the circle. On the other hand, the distance between the center of a circle and the point of contact is the radius of the circle. Hence, the radius must be the shortest distance between the center of the circle and the tangent, since all other points on the tangent lie outside of the circle. Since the radius is the shortest line segment connecting the center of the circle to the tangent, it must be perpendicular to the tangent. This proves the theorem.

In our first example, we will use this theorem to find an unknown length in a diagram involving a circle and a tangent.

### Example 1: Finding the Length of a Right Triangle’s Side given the Other Two Sides’ Lengths Using the Tangents’ Properties

Line is tangent to a circle of center at the point . Given that , , what is ?

The length, , we are looking for is a side length in triangle , so we begin by determining an angle in this triangle. We can determine by recalling that a tangent to a circle is perpendicular to the radius at the point of contact.

We are given that is tangent to the circle centered at at , and we can see that is a radius of the circle centered at , intersecting with the tangent at the point of contact. This tells us that is a right angle; hence, is a right triangle.

The length of the line , which we are looking for, is the hypotenuse of a right triangle. Using the Pythagorean theorem, we can write

We are given the length of one side of this triangle, . The remaining side is a diameter of the circle, which is twice the radius. Since we are given the radius , the diameter must be cm. This leads to . Substituting these values into the equation above, we have

Hence, .

In the previous example, we only needed the fact that a tangent and a radius are perpendicular to find a missing length. In more complex geometry problems, we may need to use more than one geometric property or theorem to find missing lengths or angles. Our next example will additionally require recalling the property of the perpendicular bisector of a chord.

### Example 2: Calculating the Perimeter of a Composite Figure Using the Chords’ Properties and the Tangents’ Properties

In the figure below, is the center of the circle, , , , and is a tangent. Find the perimeter of the figure .

Recall that a perimeter of a quadrilateral is the sum of its side lengths. Let us begin by adding the given lengths to the diagram and also highlighting the perimeter we want to compute.

We can see that two of the lengths, and , included in the perimeter are already provided. Hence, we need to obtain the lengths and .

We will start with the length . From the diagram, we can see that is the midpoint of the chord . Recall that the perpendicular bisector of a chord passes through the center of the circle. Since bisects the chord and passes through the center of the circle, it must be the perpendicular bisector of the chord. This tells us that is a right angle; hence, is a right triangle. Applying the Pythagorean theorem to this triangle, we can write

Substituting the provided lengths and into this equation and solving yields

This gives us .

Next, we will find the length . Recall that a tangent to a circle is perpendicular to the radius at the point of contact. In the given diagram, is a tangent to the circle intersecting with the radius , so must be a right angle. We can then apply the Pythagorean theorem to the right triangle to write

We are given and , so

This gives us . Then, the perimeter of the figure is

In previous examples, we applied the theorem stating that a tangent to a circle is perpendicular to the radius at the point of contact to find missing lengths. Another application of this theorem concerns the relationship between the lengths of two tangents from one point.

### Theorem: Lengths of Two Tangents from an Exterior Point

Given an exterior point to a circle, the lengths of two tangents from the point to the circle are equal.

To prove this theorem, let us consider a diagram where is the center of a circle, is an exterior point, and and are two tangents to the circle with points of contact and .

We know that the tangents intersect orthogonally with the radii, which tells us and are right angles as indicated in the diagram. The lengths of and can be obtained by using the Pythagorean theorem on the two right triangles and :

Since the right-hand sides of both equations are the same, we can equate the left-hand sides of both equations to obtain

We also know that the sides and have equal lengths, since they are radii of the same circle. Hence, the terms and in the equation above cancel each other out, leading to

This gives us , which means that the lengths of the two tangents and are equal, as claimed in the theorem.

Let us consider an example where we use this theorem to find missing lengths in a diagram involving two tangents to a circle from an exterior point.

### Example 3: Finding the Lengths of Two Line Segments Using Properties of Circle Tangents

Determine and , rounding to the nearest hundredth.

In the given diagram, and are two tangents from an exterior point to the circle centered at . Recall that the lengths of two tangents from an exterior point to a circle are equal. Hence, the lengths of these tangents must be equal. Since we are given , we must also have .

Next, let us consider . We recall that a tangent to a circle is perpendicular to the radius at the point of contact. In the diagram, is a tangent to the circle centered at , and is a radius of the circle. Hence, must be a right angle. This tells us that is a right triangle, where is the hypotenuse of this right triangle. Using the Pythagorean theorem, we can write

Substituting and into this equation gives

Rounding to the nearest hundredth, and are 12.29 cm and 10.73 cm respectively.

In the next example, we will identify unknown constants in a diagram involving two circles sharing tangents.

### Example 4: Finding the Length of a Tangent to a Circle by Solving Two Linear Equations

The two circles centered at and are touching externally. is a common tangent to them at and , respectively, and is a common tangent to them at and respectively. Given that and , find and .

Recall that the lengths of two tangents from an exterior point to a circle are equal. For the circle centered at , the lines and are tangents to this circle at points and respectively. Hence, we must have , which means . We can write

Next, consider the circle centered at . The lines and are tangents to this circle at points and respectively. Hence, we must have . We are given and we know that , so

Since , we know that . Since , we can substitute the known lengths to write

We are also given that ; we can write

Thus, we have , .

In the previous two examples, we applied the fact that two tangents from the same point to a circle have equal lengths to find missing lengths. This property of two tangents leads to a number of interesting geometric theorems. In particular, we will consider two theorems which follow from this property.

### Theorem: Bisector for an Angle Formed by Two Tangents and the Central Angle Formed by Two Radii Intersecting with the Tangents

The line connecting an exterior point to a circle to the center of the circle bisects both the angle formed by two tangents from the point to the circle and the central angle formed by the two radii intersecting with the tangents.

To prove this theorem, we consider the following diagram.

In the diagram above, the lines and are the two tangents from the exterior point to the circle centered at . We know that a tangent intersects with a radius perpendicularly at the point of contact, so and are right angles as indicated in the diagram. We also know that the lengths of two tangents emanating from the same exterior point are equal, which tells us as indicated. Finally, we know that radii of the same circle have equal lengths; hence, as indicated.

We can conclude that and are congruent by using the side-angle-side criterion for congruence. This gives us the congruence of the corresponding angles , which means that is the bisector of . Similarly, the congruence of these triangles also tell us , which tells us that is the bisector of . This proves the theorem above.

In the next example, we will apply the above theorem together with the two theorems that we have previously introduced to find the measure of an angle.

### Example 5: Finding the Measure of an Angle Using the Properties of Tangents to the Circle

Given that , where and are tangent to the circle at and , find .

We are given that and are tangent to the circle at and . We recall that the two tangents emanating from the same point to a circle have equal lengths, which gives us . Let us add an indication of this fact as well as the provided angle to the diagram.

Recall also that a tangent to a circle is perpendicular to the radius at the point of contact. In particular, this means that is the right angle. We can compute

In the diagram above, we can see that is an isosceles triangle because . This means that

Since the interior angles of a triangle add to , we can write

Substituting in the two angles that we have obtained,

Finally, to obtain , we need to recall that the line joining the exterior point and the center of the circle is the bisector of the angle between two tangents emanating from the point. This means

Hence, .

In the previous example, we applied the theorem that states that the line connecting the center of a circle to an exterior point bisects both the angle formed by the two tangents from the exterior point and the central angle formed by the two radii to the points of contact. Another application of this theorem can lead to the following perpendicular bisector theorem.

### Theorem: Perpendicular Bisector for the Chord Joining the Points of Contact of Two Tangents Emanating from an Exterior Point

Given an exterior point to a circle and two tangents from the point to the circle, the line joining the exterior point and the center of the circle is the perpendicular bisector of the chord between the points of contact of the two tangents.

Consider the following diagram, where is the center of a circle and and are tangents at points of contact and respectively.

From the previous theorem, we know that is the bisector of the angle , which tells us that as indicated in the diagram. We also know that since these are radii of the same circle. Additionally, the side is shared by the two triangles and . By the side-angle-side congruence criterion, the triangles and are congruent.

In particular, this means that , which means that is the midpoint of the chord . Also, since and these angles add to be , both of these angles must be right angles. This means that bisects the chord perpendicularly. This proves the theorem.

We now turn our attention to applications of tangents for a circle in problems involving polygons.

### Definition: Inscribed Circles and Polygons

A circle is inscribed in a polygon if each side of the polygon is a tangent to the circle.

A polygon is inscribed in a circle if the polygon lies inside the circle and all vertices of the polygon lie on the circle.

In the next example, we will find the area of a triangle inscribed in a circle with another smaller circle that is inscribed in the triangle.

### Example 6: Finding the Area of a Triangle given the Radii of Its Circumcircle and Inscribed Circle

The concentric circles shown have radii 16 cm and 8 cm. Find the area of the triangle rounded to two decimal places.

In the given diagram, the lines , , and are tangent to the smaller circle. Recall that the line joining the exterior point and the center of the circle is the bisector of the angle between the two tangents emanating from the point. Hence, lines , , and are the angle bisectors of , , and respectively. We add these lines to the diagram.

We also recall that a tangent to a circle is perpendicular to the radius at the point of contact. Thus, the radii , , and of the smaller circle intersect perpendicularly with the tangents , , and respectively. We now add these lines to the diagram.

In the diagram above, our triangle is split into six smaller right triangles. We claim that all six right triangles are congruent.

To prove the congruence, let us first consider the angles. Since we know that , , and are the angle bisectors of , , and , we know that the pair of angles at each vertex , , and have the same measure. Additionally, we note that , , and are radii of the larger circle; hence, they have equal lengths. This tells us that the triangles , , and are isosceles, meaning that the pair of angles away from the center vertex in each of these triangles are equal. Together, this tells us that all six smaller angles at the vertices , , and have equal measure. Then each of the six smaller right triangles shares this angle.

Since interior angles of a triangle must sum to , the third angles (the angle at the center vertex ) of the six right triangles must also have equal measure. Finally, reusing the fact that , , and have equal lengths, all six smaller right triangles satisfy the angle-side-angle criterion for congruence.

In particular, this means that the area of the triangle is six times the area of the triangle , for instance. Let us find the area of the right triangle . Since is a radius of the smaller circle, we know that . We need to find to find the area of this triangle. Applying the Pythagorean theorem to this triangle, we can write

We know that and is a radius of the larger circle; hence, . Substituting these values,

Recalling that the area of a triangle is one-half the lengths of the base times the height, the area of the triangle is

Multiplying this area by 6, we obtain the area of triangle rounded to the nearest hundredth; that is,

Let us finish by recapping a few important concepts from this explainer.

### Key Points

• Any tangent to a circle is perpendicular to the radius at the point of contact.
• Given an exterior point to a circle, the lengths of two tangents from the point to the circle are equal.
• The line connecting an exterior point to a circle to the center of the circle bisect both the angle formed by two tangents from the point to the circle and the central angle formed by the two radii intersecting with the tangents.
• Given an exterior point to a circle and two tangents from the point to the circle, the line joining the exterior point and the center of the circle is the perpendicular bisector of the chord between the points of contact of the two tangents.
• A circle is inscribed in a polygon if each side of the polygon is a tangent to the circle. A polygon is inscribed in a circle if the polygon lies inside the circle and all vertices of the polygon lie on the circle.