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.
