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 𝐴𝐶 is tangent to a circle of center 𝑀 at the point 𝐴. Given that 𝐵𝑀=55cm, 𝐴𝐶=96cm, what is 𝐵𝐶?

Answer

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

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

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

We are given the length of one side of this triangle, 𝐴𝐶=96cm. The remaining side 𝐴𝐵 is a diameter of the circle, which is twice the radius. Since we are given the radius 𝐵𝑀=55cm, the diameter must be 55×2=110 cm. This leads to 𝐴𝐵=110cm. Substituting these values into the equation above, we have 96+110=𝐵𝐶𝐵𝐶=96+110=146.

Hence, 𝐵𝐶=146cm.

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, 𝑀 is the center of the circle, 𝑀𝐵=15cm, 𝐴𝐵=20cm, 𝑀𝐶=9cm, and 𝐴𝐵 is a tangent. Find the perimeter of the figure 𝐴𝐵𝐶𝑀.

Answer

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, 𝑀𝐶 and 𝐴𝐵, included in the perimeter are already provided. Hence, we need to obtain the lengths 𝑀𝐴 and 𝐵𝐶.

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

Substituting the provided lengths 𝑀𝐶=9cm and 𝑀𝐵=15cm into this equation and solving yields 𝐵𝐶+9=15𝐵𝐶=159=144𝐵𝐶=144=12.

This gives us 𝐵𝐶=12cm.

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

We are given 𝐴𝐵=20cm and 𝑀𝐵=15cm, so 15+20=𝑀𝐴𝑀𝐴=15+20=25.

This gives us 𝑀𝐴=25cm. Then, the perimeter of the figure 𝐴𝐵𝐶𝑀 is 𝑀𝐴+𝐴𝐵+𝐵𝐶+𝑀𝐶=25+20+12+9=66.cm

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 𝑀 is the center of a circle, 𝐴 is an exterior point, and 𝐴𝐵 and 𝐴𝐶 are two tangents to the circle with points of contact 𝐵 and 𝐶.

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

Since the right-hand sides of both equations are the same, we can equate the left-hand sides of both equations to obtain 𝐴𝐶+𝑀𝐶=𝐴𝐵+𝑀𝐵.

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

This gives us 𝐴𝐶=𝐴𝐵, which means that the lengths of the two tangents 𝐴𝐶 and 𝐴𝐵 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 𝐴𝑀 and 𝐴𝐵, rounding to the nearest hundredth.

Answer

In the given diagram, 𝐴𝐶 and 𝐴𝐵 are two tangents from an exterior point 𝐴 to the circle centered at 𝑀. 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 𝐴𝐶=10.73cm, we must also have 𝐴𝐵=10.73cm.

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

Substituting 𝐴𝐵=10.73cm and 𝐵𝑀=6cm into this equation gives 10.73+6=𝐴𝑀𝐴𝑀=10.73+6=12.2936.

Rounding to the nearest hundredth, 𝐴𝑀 and 𝐴𝐵 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 𝑀 and 𝑁 are touching externally. 𝐹𝐴 is a common tangent to them at 𝐴 and 𝐵, respectively, and 𝐹𝐶 is a common tangent to them at 𝐶 and 𝐷 respectively. Given that 𝐴𝐵=11.01cm and 𝐶𝐷=(𝑦11.01)cm, find 𝑥 and 𝑦.

Answer

Recall that the lengths of two tangents from an exterior point to a circle are equal. For the circle centered at 𝑁, the lines 𝐹𝐴 and 𝐹𝐶 are tangents to this circle at points 𝐵 and 𝐷 respectively. Hence, we must have 𝐹𝐵=𝐹𝐷, which means 𝐹𝐵=12.31cm. We can write 𝑥2=12.31,𝑥=14.31.whichleadsto

Next, consider the circle centered at 𝑀. The lines 𝐹𝐴 and 𝐹𝐶 are tangents to this circle at points 𝐴 and 𝐶 respectively. Hence, we must have 𝐹𝐴=𝐹𝐶. We are given 𝐴𝐵=11.01cm and we know that 𝐹𝐵=12.31cm, so 𝐹𝐴=𝐹𝐵+𝐴𝐵=12.31+11.01=23.32.cm

Since 𝐹𝐴=𝐹𝐶, we know that 𝐹𝐶=23.32cm. Since 𝐹𝐶=𝐹𝐷+𝐶𝐷, we can substitute the known lengths to write 23.32=12.31+𝐶𝐷.

This leads to 𝐶𝐷=23.3212.31=11.01.cm

We are also given that 𝐶𝐷=(𝑦11.01)cm; we can write 𝑦11.01=11.01,𝑦=22.02.whichleadsto

Thus, we have 𝑥=14.31, 𝑦=22.02.

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 𝐴𝐵 and 𝐴𝐶 are the two tangents from the exterior point 𝐴 to the circle centered at 𝑀. We know that a tangent intersects with a radius perpendicularly at the point of contact, so 𝐴𝐵𝑀 and 𝐴𝐶𝑀 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 𝐴𝐵=𝐴𝐶 as indicated. Finally, we know that radii of the same circle have equal lengths; hence, 𝐵𝑀=𝐶𝑀 as indicated.

We can conclude that 𝐴𝐵𝑀 and 𝐴𝐶𝑀 are congruent by using the side-angle-side criterion for congruence. This gives us the congruence of the corresponding angles 𝑚𝐵𝑀𝐴=𝑚𝐶𝑀𝐴, which means that 𝑀𝐴 is the bisector of 𝐵𝑀𝐶. Similarly, the congruence of these triangles also tell us 𝑚𝐵𝐴𝑀=𝑚𝐶𝐴𝑀, which tells us that 𝑀𝐴 is the bisector of 𝐵𝐴𝐶. 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 𝑚𝑀𝐶𝐵=49, where 𝐴𝐵 and 𝐴𝐶 are tangent to the circle at 𝐵 and 𝐶, find 𝑚𝐵𝐴𝑀.

Answer

We are given that 𝐴𝐵 and 𝐴𝐶 are tangent to the circle at 𝐵 and 𝐶. We recall that the two tangents emanating from the same point to a circle have equal lengths, which gives us 𝐴𝐵=𝐴𝐶. 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 𝐴𝐶𝑀 is the right angle. We can compute 𝑚𝐴𝐶𝐵=𝑚𝐴𝐶𝑀𝑚𝑀𝐶𝐵=9049=41.

In the diagram above, we can see that 𝐴𝐵𝐶 is an isosceles triangle because 𝐴𝐵=𝐴𝐶. This means that 𝑚𝐴𝐵𝐶=𝑚𝐴𝐶𝐵=41.

Since the interior angles of a triangle add to 180, we can write 𝑚𝐵𝐴𝐶+𝑚𝐴𝐵𝐶+𝑚𝐴𝐶𝐵=180.

Substituting in the two angles that we have obtained, 𝑚𝐵𝐴𝐶+41+41=180.

This leads to 𝑚𝐵𝐴𝐶=98.

Finally, to obtain 𝑚𝐵𝐴𝑀, 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 𝑚𝐵𝐴𝑀=12𝑚𝐵𝐴𝐶=12×98=49.

Hence, 𝑚𝐵𝐴𝑀=49.

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 𝑀 is the center of a circle and 𝐴𝐵 and 𝐴𝐶 are tangents at points of contact 𝐵 and 𝐶 respectively.

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

In particular, this means that 𝐵𝐷=𝐶𝐷, which means that 𝐷 is the midpoint of the chord 𝐵𝐶. Also, since 𝑚𝐵𝐷𝑀=𝑚𝐶𝐷𝑀 and these angles add to be 180, both of these angles must be right angles. This means that 𝑀𝐴 bisects the chord 𝐵𝐶 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.

Answer

In the given diagram, the lines 𝐵𝐶, 𝐴𝐶, and 𝐴𝐵 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 𝑀𝐴, 𝑀𝐵, and 𝑀𝐶 are the angle bisectors of 𝐶𝐴𝐵, 𝐴𝐵𝐶, and 𝐵𝐶𝐴 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 𝑀𝑋, 𝑀𝑌, and 𝑀𝑍 of the smaller circle intersect perpendicularly with the tangents 𝐴𝐵, 𝐵𝐶, and 𝐶𝐴 respectively. We now add these lines to the diagram.

In the diagram above, our triangle 𝐴𝐵𝐶 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 𝑀𝐴, 𝑀𝐵, and 𝑀𝐶 are the angle bisectors of 𝐶𝐴𝐵, 𝐴𝐵𝐶, and 𝐵𝐶𝐴, we know that the pair of angles at each vertex 𝐴, 𝐵, and 𝐶 have the same measure. Additionally, we note that 𝑀𝐴, 𝑀𝐵, and 𝑀𝐶 are radii of the larger circle; hence, they have equal lengths. This tells us that the triangles 𝑀𝐴𝐶, 𝑀𝐴𝐵, and 𝑀𝐵𝐶 are isosceles, meaning that the pair of angles away from the center vertex 𝑀 in each of these triangles are equal. Together, this tells us that all six smaller angles at the vertices 𝐴, 𝐵, and 𝐶 have equal measure. Then each of the six smaller right triangles shares this angle.

Since interior angles of a triangle must sum to 180, the third angles (the angle at the center vertex 𝑀) of the six right triangles must also have equal measure. Finally, reusing the fact that 𝑀𝐴, 𝑀𝐵, and 𝑀𝐶 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 𝐴𝐵𝐶 is six times the area of the triangle 𝐴𝑀𝑋, for instance. Let us find the area of the right triangle 𝐴𝑀𝑋. Since 𝑀𝑋 is a radius of the smaller circle, we know that 𝑀𝑋=8cm. We need to find 𝐴𝑋 to find the area of this triangle. Applying the Pythagorean theorem to this triangle, we can write 𝐴𝑋+𝑀𝑋=𝐴𝑀.

We know that 𝑀𝑋=8cm and 𝐴𝑀 is a radius of the larger circle; hence, 𝐴𝑀=16cm. Substituting these values, 𝐴𝑋+8=16𝐴𝑋=168=192𝐴𝑋=192.cm

Recalling that the area of a triangle is one-half the lengths of the base times the height, the area of the triangle 𝐴𝑀𝑋 is 12×𝐴𝑋×𝑀𝑋=12×192×8=4192.cm

Multiplying this area by 6, we obtain the area of triangle 𝐴𝐵𝐶 rounded to the nearest hundredth; that is, 6×4192=24192=332.55.cm

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.

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