Lesson Video: Tangents of a Circle | Nagwa Lesson Video: Tangents of a Circle | Nagwa

Lesson Video: Tangents of a Circle Mathematics • Third Year of Preparatory School

Join Nagwa Classes

Attend live Mathematics sessions on Nagwa Classes to learn more about this topic from an expert teacher!

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

14:50

Video Transcript

In this video, we will learn how to use the properties of tangents of circles to find missing angles or side lengths. Let’s begin by recalling what we mean by a tangent.

A tangent to a circle is a straight line that intersects the circle at a single point. It doesn’t pass inside the circle but instead just touches the circle at its circumference. The first key property we’re going to consider is this. Any tangent to a circle is perpendicular to the radius at the point of contact. This just means that any tangent we draw to a circle forms a right angle with the radius of the circle at that point. The tangent is also perpendicular to the diameter at the point of contact as this is just an extension of the radius. We won’t go into the detail of the proof of this theorem here, but it relies on the fact that the shortest distance between a line and a point is the perpendicular distance between the two objects.

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

Line 𝐴𝐶 is tangent to a circle of center 𝑀 at the point 𝐴. Given that 𝐵𝑀 equals 55 centimeters and 𝐴𝐶 equals 96 centimeters, what is 𝐵𝐶?

Let’s begin by adding the information in the question to the diagram. 𝐵𝑀 is 55 centimeters and 𝐴𝐶 is 96 centimeters. We’re then asked to find the length of the line segment 𝐵𝐶. Now, we can see that the line segments 𝐴𝐵, 𝐴𝐶, and 𝐵𝐶 form a triangle. We can also deduce that the length of the side 𝐴𝐵 of this triangle is 110 centimeters. That’s because 𝑀𝐵 and 𝑀𝐴 are each radii of the circle, and so they’re of equal length. We now know two of the side lengths in triangle 𝐴𝐵𝐶, and we wish to calculate the third. This suggests we may want to use the Pythagorean theorem, but we can only do this if the triangle we’re working with is a right triangle.

Remember that the line 𝐴𝐶 is a tangent to the circle, and so we can recall a key property. Any tangent to a circle is perpendicular to the radius at the point of contact. So the line 𝐴𝐶 is perpendicular to the radius at the point 𝐴. That’s the line segment 𝐴𝑀. And by extension, it’s also perpendicular to the line segment 𝐴𝐵. So triangle 𝐴𝐵𝐶 is right angled, and we can therefore apply the Pythagorean theorem. 𝐵𝐶 is the hypotenuse of this triangle, so we have that 𝐴𝐵 squared plus 𝐴𝐶 squared is equal to 𝐵𝐶 squared. Substituting the lengths, 110 squared plus 96 squared is equal to 𝐵𝐶 squared. Evaluating, we find that 𝐵𝐶 squared is equal to 21,316. 𝐵𝐶 is the square root of this value, which is 146.

So by recalling that any tangent to a circle is perpendicular to the radius at the point of contact and then applying the Pythagorean theorem, we found that 𝐵𝐶 is 146 centimeters.

The second key property relating to tangents of circles is this. Given an exterior point to a circle, the lengths of two tangents from this point to the circle are equal. So in the figure, we have a point 𝐴, which is exterior to the circle 𝑀, and then two tangents drawn from this point to the circle. And what we’re saying is that the length 𝐴𝐵 is equal to the length 𝐴𝐶. We can prove this using congruent triangles and the first property of tangents to circles. Let’s draw in the line segment 𝐴𝑀 and then consider the two triangles 𝐴𝐵𝑀 and 𝐴𝐶𝑀.

We know from the first property in this lesson that each of the tangents 𝐴𝐵 and 𝐴𝐶 are perpendicular to the radius of the circle at their point of contact. So 𝐴𝐵 is perpendicular to 𝐵𝑀 and 𝐴𝐶 is perpendicular to 𝐶𝑀. So in fact, each of the triangles 𝐴𝐵𝑀 and 𝐴𝐶𝑀 are right triangles. The line segments 𝐵𝑀 and 𝐶𝑀 are each radii of the circle, and so they are of equal length. And finally, the side 𝐴𝑀 is shared between the two triangles. So using the RHS, right-angle-hypotenuse-side, congruency condition, these two triangles are congruent. Therefore, 𝐴𝐵 and 𝐴𝐶, which are corresponding sides in these two triangles, will be the same length, and so we’ve proven this result.

We won’t consider any examples of this property here. But typical questions may include being given expressions for the lengths of two tangents drawn from a point to a circle in terms of an unknown and being required to find the value of that unknown. Of course, we would do this by recognizing that the two tangent segments are equal in length, equating the two expressions, and then solving the resulting equation to find the value of the unknown.

We’ll now consider two corollaries derived from this theorem. The first is this. 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.

Let’s understand what this means on the diagram. Here we have an external point, and we have two tangents drawn from this point to a circle. The angle between those two tangents is here, and then we have a line connecting the external point to the center of the circle. The theorem tells us that this line is bisecting the angle between the two tangents. We also have two radii drawn from where each tangent intersects the circle to its center. The angle between those two radii is here, and once again the theorem is telling us that the line from the exterior point to the center of the circle bisects this angle.

The second corollary is this. 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. So on the diagram, here we have the external point, and then we have the two tangents drawn from this point to the circle. Here’s the line that joins the exterior point to the center of the circle. And we’re told that this line is the perpendicular bisector of the chord which connects the points of contact of the two tangents.

Let’s now consider an example in which we apply one of these results.

Given that the measure of angle 𝑀𝐴𝐶 equals 36 degrees, determine the measure of angle 𝐵𝐴𝑀 and the measure of angle 𝐴𝑀𝐶.

We’ll begin by adding the information in the question to the diagram. The measure of angle 𝑀𝐴𝐶 is 36 degrees. We need to find the measures of angles 𝐵𝐴𝑀 and 𝐴𝑀𝐶. First, we should know that 𝐴 is a point exterior to the circle. The line segments 𝐴𝐶 and 𝐴𝐵 are tangents of the circle, and the line segment 𝐴𝑀 is the line segment connecting this exterior point to the center of the circle. We can recall that the line connecting an exterior point to the center of a circle bisects the angle formed by two tangents from that point to the circle. So at point 𝐴, the measures of angles 𝑀𝐴𝐵 and 𝑀𝐴𝐶 are equal. We know the measure of angle 𝑀𝐴𝐶 is 36 degrees, and so the measure of angle 𝑀𝐴𝐵 or angle 𝐵𝐴𝑀 is also 36 degrees.

We’ll now find the measure of angle 𝐴𝑀𝐶, and to do this, we’ll consider triangle 𝐴𝑀𝐶. 𝐴𝐶 is a tangent of the circle and 𝑀𝐶 is a radius. We recall that any tangent to a circle is perpendicular to the radius at the point of contact, and so the angle 𝑀𝐶𝐴 is a right angle. Using the fact that the angle sum in any triangle is 180 degrees, we find the measure of angle 𝐴𝑀𝐶 by subtracting the measures of the other two angles in triangle 𝐴𝑀𝐶 from 180 degrees, which gives 54 degrees. The measure of angle 𝐵𝐴𝑀 is 36 degrees, and the measure of angle 𝐴𝑀𝐶 is 54 degrees.

We’ll now consider the applications of tangents to a circle in problems involving polygons. We’ll consider first the definitions of inscribed circles and inscribed polygons. A circle is inscribed in a polygon if each side of the polygon is a tangent to the circle. So in the figure here, we have a circle inscribed within a triangle, and each side of the triangle is a tangent to the circle. On the other hand, a polygon is inscribed in a circle if the polygon lies inside the circle and all vertices of the polygon lie on the circle. So in this figure, we have a quadrilateral inscribed in a circle. It’s inside the circle, and each of its vertices lies on the circle’s circumference.

We’ll now consider one final example involving inscribed circles and inscribed polygons.

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

Looking at the diagram, we note first that the smaller circle is inscribed within triangle 𝐴𝐵𝐶 as each of the sides of triangle 𝐴𝐵𝐶 are tangent to the smaller circle. Triangle 𝐴𝐵𝐶 is itself inscribed within the larger circle as each of its vertices lies on the circumference of the larger circle. So we have a circle inscribed in a polygon, in this case a triangle, which is itself inscribed in a circle. And we need to find the area of this triangle.

We’ll begin by sketching in the radii of the larger circle, the line segments 𝑀𝐴, 𝑀𝐵, and 𝑀𝐶, each of which are of length 16 centimeters. We can also sketch in some radii of the smaller circle, the line segments 𝑀𝑋, 𝑀𝑌, and 𝑀𝑍, each of which are of length eight centimeters. Now, in doing so, we have divided triangle 𝐴𝐵𝐶 into six smaller triangles. It will be really useful if these triangles are congruent. So let’s see if we can prove it. Each triangle has one side of length eight centimeters and one side of length 16 centimeters. As the smaller circle is inscribed within triangle 𝐴𝐵𝐶, each side of 𝐴𝐵𝐶 is tangent to the smaller circle.

We recall that any tangent to a circle is perpendicular to the radius at the point of contact. So the angles 𝐶𝑌𝑀, 𝐵𝑌𝑀, 𝐵𝑋𝑀, 𝐴𝑋𝑀, 𝐴𝑍𝑀, and 𝐶𝑍𝑀 are all right angles. We have therefore shown that these triangles are all right triangles, with the same length of hypotenuse and another side the same length. So by the RHS, right-angle-hypotenuse-side, congruency condition, these six triangles are congruent to one another.

Let’s consider just one of these triangles then, triangle 𝑀𝐶𝑌. This is a right triangle, so its area can be found using the formula base multiplied by perpendicular height over two. That’s 𝐶𝑌 multiplied by 𝑀𝑌 over two. We know the length of 𝑀𝑌. It’s the radius of the smaller circle, so it’s eight centimeters. And to find the length of 𝐶𝑌, we can apply the Pythagorean theorem. 𝐶𝑌 squared plus eight squared is equal to 16 squared. Evaluating, we have 𝐶𝑌 squared is equal to 192, and so 𝐶𝑌 is equal to the square root of 192 or in simplified form eight root three. The area of triangle 𝑀𝐶𝑌 then is eight root three multiplied by eight over two, that’s 64 root three over two, which is equal to 32 root three.

To find the area of triangle 𝐴𝐵𝐶, we need to multiply this value by six as there were six congruent triangles. That gives 192 root three or as a decimal 332.5537 continuing. To two decimal places then, the area of triangle 𝐴𝐵𝐶, which inscribes the smaller circle and is inscribed in the larger circle, is 332.55 square centimeters.

Let’s now summarize the key points from this video. Any tangent to a circle is perpendicular to the radius at the point of contact. Tangents drawn from the same exterior point to the circumference of a circle are equal in length. The line connecting an exterior point to the center of a circle bisects both the angle formed by two tangents from that point to the circle and the central angle formed by the two radii intersecting the tangents.

Given an exterior point to a circle and two tangents drawn from that 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 tangents. A circle is inscribed in a polygon if each side of the polygon is a tangent to the circle. And finally, a polygon is inscribed in a circle if the polygon is inside the circle and each of the polygon’s vertices lie on the circle’s circumference.

Join Nagwa Classes

Attend live sessions on Nagwa Classes to boost your learning with guidance and advice from an expert teacher!

  • Interactive Sessions
  • Chat & Messaging
  • Realistic Exam Questions

Nagwa uses cookies to ensure you get the best experience on our website. Learn more about our Privacy Policy