Video: Tangents of a Circle

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

16:09

Video Transcript

In this video, we will learn how to use the properties of tangents of circles to find the missing angles or side lengths. This is part of a wider subject with circle theorems, which focuses on properties of angles formed inside circles by chords, tangents, and radii. In this lesson, we’ll be specifically focusing on properties of the angles and lengths made by tangents drawn from exterior points to the circumference of a circle. You should, however, be familiar with general angle properties such as the sum of angles on a straight line and the sum of angles in a triangle.

Remember, first of all, that a tangent to the circle is a straight line that intersects the circle at a single point. It doesn’t pass inside the circle, but instead just meets the circle at a point on its circumference.

The first key property we’re going to consider is this: a tangent to a circle is perpendicular to the radius at the point of contact. What this means is that if we draw in the radius of the circle from the point on the circumference where the tangent touches the circle, then the angle between the tangent and the radius will be a right angle. Now, of course, it’s also true that the tangent will be perpendicular to the diameter of the circle at this point, as this is just a continuation of the radius. But it’s the radius that we tend to use when quoting this result. Now, the proof of this requires some of the other circle theorems, including one called the alternate segment theorem. So, we won’t go into it here. However, we will see a proof of another key property later in this video. So, we’ll still get a flavor of how to prove these theorems.

Let’s now consider some examples using this first property.

Given that the line segment 𝐴𝐵 is a tangent to the circle 𝑀, and the measure of angle 𝑀𝐵𝐹 is 123 degrees, determine the measure of angle 𝐴𝑀𝐵.

Angle 𝐴𝑀𝐵 is the angle formed when we travel from 𝐴 to 𝑀 to 𝐵. So, it’s the angle marked in orange on the figure. Angle 𝑀𝐵𝐹 is the angle made when we travel from 𝑀 to 𝐵 to 𝐹. So, it’s the angle now labelled in pink, and its measure is 123 degrees. We can see that the angle we’re looking to find — angle 𝐴𝑀𝐵 — is contained within a triangle. If we can work out the other two angles in this triangle, then we can use the fact that the angle sum in any triangle is 180 degrees to find the angle we’re looking for.

Firstly, let’s consider the angle 𝑀𝐵𝐴. One of our most basic angle facts is that angles on a straight line sum to 180 degrees. And this angle is on a straight line with the angle we’ve already marked as being 123 degrees. So, we can say that angle 𝑀𝐵𝐹 plus angle 𝑀𝐵𝐴 equals 180 degrees. As stated, we already know the measure of angle 𝑀𝐵𝐹. So, we can substitute this value. And we now have an equation we can solve to find the measure of angle 𝑀𝐵𝐴. We need to subtract 123 from each side of this equation. Doing so and we find that angle 𝑀𝐵𝐴 is 57 degrees.

So, we found one of the angles in triangle 𝑀𝐵𝐴. Can we find another one? What about angle 𝑀𝐴𝐵? Well, this is the angle formed where a tangent to the circle — that’s the line 𝐴𝐵 — meets the radius of the circle 𝐴𝑀. And we know that a tangent to a circle is perpendicular to the radius at the point of contact. So, we know that angle 𝑀𝐴𝐵 is 90 degrees. It’s a right angle. We’ve, therefore, found two of the angles in triangle 𝐴𝐵𝑀. And using the angle sum in a triangle, we can find the third.

We have the equation angle 𝐴𝑀𝐵 plus 90 degrees plus 57 degrees equals 180 degrees. 90 plus 57 is 147. And subtracting 147 degrees from each side of the equation gives angle 𝐴𝑀𝐵 equals 33 degrees. So, by using two basic angles facts as well as the key result that a tangent to a circle is perpendicular to the radius at the point of contact, we found that the measure of angle 𝐴𝑀𝐵 is 33 degrees.

Let’s now consider a second application of this important result.

Given that the line segment 𝐴𝐵 is tangent to the circle 𝑀 at 𝐴, 𝐴𝑀 equals 8.6 centimeters, and 𝑀𝐵 equals 12.3 centimeters, find the length of the line segment 𝐴𝐵, and round the result to the nearest tenth.

Let’s begin by adding the information given in the question onto the diagram. 𝐴𝑀 is 8.6 centimeters. That’s this length here. 𝑀𝐵 is 12.3 centimeters. That’s this length here. And the length we’re looking to find is the length of the line segment 𝐴𝐵. Now, we notice that we have a triangle, triangle 𝐴𝑀𝐵, in which we know the lengths of two of the sides. Your first thought then may be that we could apply the Pythagorean theorem. But remember, the Pythagorean theorem is only valid in right triangles. So, we need to consider whether the triangle 𝐴𝑀𝐵 is a right triangle.

The other key piece of information given in the question is that the line segment 𝐴𝐵 is tangent to the circle 𝑀 at 𝐴. A key property about tangents to circles is that a tangent to a circle is perpendicular to the radius at the point of contact. So, the line segment 𝐴𝐵 is perpendicular to the radius 𝐴𝑀. And we, therefore, have a right angle at 𝐴 in our triangle 𝐴𝑀𝐵. We do indeed have a right triangle. And so, we can apply the Pythagorean theorem to calculate the length of the third side.

The Pythagorean theorem tells us that in a right triangle, the sum of the squares of the two shorter sides which we can think of as 𝑎 and 𝑏 is equal to the square of the longest side of the triangle, which we can think of as 𝑐. Remember, the longest side or hypotenuse is always the side directly opposite the right angle. So, in this case, that’s the side 𝑀𝐵. Substituting 𝐴𝐵 and 8.6 for the two shorter sides of the triangle and 12.3 for the longest side or hypotenuse, we have the equation 𝐴𝐵 squared plus 8.6 squared equals 12.3 squared. We can evaluate 8.6 squared and 12.3 squared and then subtract 73.96 — that’s 8.6 squared — from each side, giving 𝐴𝐵 squared equals 77.33.

We solve this equation by square rooting. And we’re only going to take the positive value here as 𝐴𝐵 has a physical meaning as the length of a side in this triangle. Evaluating this square root on a calculator, we find that 𝐴𝐵 is equal to 8.79374. Remember, though, that we were asked to around the result to the nearest tenth. So, as there is a nine in the hundreds column, we round up, giving 𝐴𝐵 equals 8.8 centimeters.

So in this problem, by applying the key property that a tangent to a circle is perpendicular to the radius at the point of contact, we were able to deduce that triangle 𝑀𝐴𝐵 was a right triangle. And hence, we can apply the Pythagorean theorem to calculate the length of its third side. We found that the length of 𝐴𝐵 to the nearest tenth is 8.8 centimeters.

Let’s now look at one final example of how we can apply this first property.

Given that the line segment 𝐴𝐵 is a tangent to the circle 𝑀, and the measure of angle 𝐴𝐵𝑀 is 49 degrees, determine the measure of angle 𝐴𝐷𝐵.

Angle 𝐴𝐷𝐵 is the angle formed when we travel from 𝐴 to 𝐷 to 𝐵. So, it’s the angle marked in orange on the diagram. Angle 𝐴𝐵𝑀 is the angle formed when we travel from 𝐴 to 𝐵 to 𝑀. It’s the angle now marked in pink on the diagram with its measure of 49 degrees. From the information given, we aren’t able to calculate angle 𝐴𝐷𝐵 directly. We’re going to need to find the measures of some other angles in the figure first. The other key piece of information given in the question, though, is that the line 𝐴𝐵 is a tangent to the circle 𝑀. And the key property about tangents of circles is that a tangent to a circle is perpendicular to the radius at the point of contact.

The point where the tangent meets the circle is point 𝐴. And the radius here is the line segment 𝐴𝑀. So, we know that the angle 𝐵𝐴𝑀 is 90 degrees. So, we now know one more angle within the figure. We still aren’t able to calculate angle 𝐴𝐷𝐵 directly. So, let’s see what other angles we could work out. We have a triangle. In fact, it’s a right triangle, triangle 𝐴𝑀𝐵. And we know two of its angles, the right angle and the angle of 49 degrees. So, using the fact that angles in a triangle sum to 180 degrees, we can calculate the third angle in this triangle.

We have that angle 𝐴𝑀𝐵 plus 90 degrees plus 49 degrees equals 180 degrees. 90 plus 49 is 139. And subtracting this from 180, we find that angle 𝐴𝑀𝐵 is 41 degrees. So, we now know another angle in our diagram. We still don’t have enough information to calculate angle 𝐴𝐷𝐵. But we can now calculate a different angle, angle 𝐴𝑀𝐷. We know that the angles on any straight line sum to 180 degrees. So, angle 𝐴𝑀𝐷 plus the angle we’ve just calculated of 41 degrees must equal 180 degrees. Angle 𝐴𝑀𝐷 is, therefore, equal to 180 degrees minus 41 degrees. That’s 139 degrees.

Now, we found almost all of the angles in the figure, but still not the one that we were looking for. The final step is to consider triangle 𝐴𝑀𝐷, in which we know one angle of 139 degrees. We need to notice that both 𝑀𝐷 and 𝑀𝐴 are radii of the circle 𝑀. And therefore, they’re of the same length. This means that triangle 𝑀𝐷𝐴 is an isosceles triangle. And it also means that angle 𝑀𝐷𝐴 will be equal to angle 𝑀𝐴𝐷. We can, therefore, find the measure of each angle by subtracting the third angle, 139 degrees, from the total angle sum in a triangle, 180 degrees, and then halving the remainder. Doing so gives each of these angles to be 20.5 degrees. Now, angle 𝑀𝐷𝐴 is in fact the same angle as angle 𝐴𝐷𝐵. They both refer to this angle here. And so, we’ve completed the problem.

By using some of the more basic facts about angles in triangles and angles in straight lines and then the key property that a tangent to a circle is perpendicular to the radius at the point of contact, we’ve found the measure of angle 𝐴𝐷𝐵 is 20.5 degrees.

So, we’ve now seen three applications of the first key property about tangents of circles. Let’s now introduce the second property. It’s this: tangents drawn to a circle from the same exterior point are equal in length.

In our diagram, the two tangents have been drawn from the exterior point 𝐴. So, the length of the line 𝐴𝐵 will be equal to 𝐴𝐶. Now, we can prove this property using congruent triangles and the first property. We’ll add a couple of lines to our diagram, firstly, the radii 𝑀𝐵 and 𝑀𝐶 and, secondly, a line connecting our exterior point 𝐴 to the center of the circle 𝑀. Now, we’re going to consider the two triangles 𝐴𝐵𝑀 and 𝐴𝐶𝑀.

We know from the first property that both tangents will be perpendicular to the radius at the point of contact. So, we know that angle 𝐴𝐵𝑀 and angle 𝐴𝐶𝑀 are each 90 degrees. But in particular, they’re equal to one another. We also know that the line segments 𝐵𝑀 and 𝐶𝑀 are each radii of the circle. And so, they will be of equal length. And the line 𝐴𝑀 is a shared side in these two triangles. In fact, it is the hypotenuse of the two triangles. We’ve shown then that these two triangles each have a right angle, they share a common hypotenuse, and they have one of the two shorter sides of the triangle also equal in length. Therefore, the two triangles are congruent by the RHS. That’s right-angle-hypotenuse-side congruency condition. If the two triangles are congruent, then the lengths of their third sides — that’s 𝐴𝐵 and 𝐴𝐶 — must be the same. And so, we’ve shown that 𝐴𝐵 is indeed equal to 𝐴𝐶 and thus proven this property.

Let’s look then at one application of this.

Find the value of 𝑥.

In the diagram, we can see that we have a circle and then two lines 𝐴𝐵 and 𝐴𝐶, each of which are tangents to the circle. The two tangents have been drawn from the same exterior point, point 𝐴. One of the key properties of tangents of circles is that tangents drawn from the same exterior point are equal in length. So, we know that the line segment 𝐴𝐵 is equal in length to the line segment 𝐴𝐶. We’ve been given the length of 𝐴𝐵. It’s 21 centimeters. And we’ve been given an expression for the length of 𝐴𝐶. It’s two 𝑥 plus five centimeters. So, equating these, we can form an equation, two 𝑥 plus five is equal to 21.

We can now solve this equation to determine the value of 𝑥. We first subtract five from each side, giving two 𝑥 equals 16, and then divide by two, giving 𝑥 equals eight. Using the key property then, the tangents drawn from the same exterior point are equal in length, we’ve found the value of 𝑥. 𝑥 is equal to eight.

This example involves solving a simple linear equation. But more complicated examples of this type may involve setting up and solving simultaneous equations. The algebra maybe more complicated, but the principles involved will be the same.

Let’s now review what we’ve seen in this lesson. We’ve introduced two key properties of tangents to circles. Firstly, a tangent to a circle is perpendicular to the radius at the point of contact. And by extension, it’s also perpendicular to the diameter of the circle at this point. Secondly, we saw and proved that tangents drawn to a circle from the same exterior point are equal in length. In this diagram, this means that 𝐴𝐵 is equal to 𝐴𝐶. We’ve also seen that we can use these properties in partnership with other angle rules and the Pythagorean theorem to find missing angles and side lengths in problems involving tangents to circles.

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