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Lesson Explainer: Circles and Triangles

In this explainer, we will learn how to identify inscribed angles in semicircles and circumcircles of triangles and find the equation of a circle given three points on the circumference.

Given three noncolinear points 𝐴, 𝐡, and 𝐢 in the plane, there is a unique triangle 𝐴𝐡𝐢 with those points as vertices.

Now, consider the perpendicular bisector 𝐿 of the line segment 𝐴𝐡. Every point on 𝐿 lies at an equal distance from point 𝐴 and point 𝐡. Similarly, every point on the bisector 𝐿 is equidistant from 𝐡 and 𝐢.

This means that the point 𝑃 where 𝐿 and 𝐿 intersect is the same distance from all three vertices, 𝐴, 𝐡, and 𝐢.

If we now trace a circle with center 𝑃 passing through the three vertices 𝐴, 𝐡, and 𝐢 we get the unique circle with these properties. This circle is called the circumcircle of the triangle.

The point 𝑃 where the three perpendicular bisectors of the sides of the triangle intersect is the center of the circumcircle and is called the triangle’s circumcenter. This is one type of triangle center; there are many others such as the orthocenter, where the three altitudes of the triangle meet, and the centroid, where the three lines that connect each vertex to the midpoint of the opposite side meet.

Be aware that the circumcenter of a triangle does not always lie inside the triangle. When △𝐴𝐡𝐢 is acute, which is to say all of its angles are smaller than a right angle, then the circumcenter lies inside △𝐴𝐡𝐢. When △𝐴𝐡𝐢 contains an angle larger than a right angle, then the circumcenter lies outside the triangle.

Now, suppose △𝐴𝐡𝐢 is a right triangle with hypotenuse 𝐡𝐢. By looking at the angles, we can see that the the perpendicular bisector of 𝐴𝐡 makes a smaller similar triangle inside △𝐴𝐡𝐢, scaled down by a factor of 12.

Similarly, the perpendicular bisector of 𝐴𝐢 cuts out a similar triangle of scale factor 12 inside △𝐴𝐡𝐢. This means that both of these lines intersect the hypotenuse at the point 𝑄 halfway along its length. Of course, the perpendicular bisector of the hypotenuse, by definition, intersects the hypotenuse at its midpoint too. Altogether, this shows that if △𝐴𝐡𝐢 is a right triangle, then its circumcenter is the midpoint of its hypotenuse.

In this situation, the hypotenuse of △𝐴𝐡𝐢 is a diameter of its circumcircle and its right angle lies on the circumference.

Recall the circle theorem, which states that the angle in a semicircle is always 90∘, or, in other words, if two points 𝐢 and 𝐡 form the diameter of a circle, then ∠𝐢𝐴𝐡 subtended at any third point 𝐴 on the circumference must be a right angle.

What we have just established is the converse to this theorem: if △𝐴𝐡𝐢 is a right triangle with hypotenuse 𝐡𝐢, then 𝐡𝐢 is a diameter of its circumcircle.

Properties: Triangles and Circumcircles

Let 𝐴, 𝐡, and 𝐢 be three noncolinear points in the plane.

  • There is a unique triangle with vertices 𝐴, 𝐡, and 𝐢.
  • The perpendicular bisectors of the sides 𝐴𝐡, 𝐡𝐢, and 𝐴𝐢 meet at a single point 𝑃, called the circumcenter of the triangle.
  • The circumcenter of the triangle is the center of the circumcircle, which is the unique circle passing through the vertices 𝐴, 𝐡, and 𝐢.
  • If △𝐴𝐡𝐢 is a right triangle, then its circumcenter lies at the midpoint of its hypotenuse and its hypotenuse is a diameter of its circumcircle.

Let us put some of these properties to use in identifying a diameter of a circle.

Example 1: Identifying the Correct Point That Makes 𝐴𝐡 A Diameter of A Circle

𝐡(9,3), 𝐢(4,8), 𝐷(8,2), 𝐸(8,8), and 𝐹(3,7) are five points on a circle.

Which of the chords 𝐢𝐷, 𝐡𝐢, and 𝐷𝐹 is a diameter?

Answer

To identify a diameter, we will use the fact that two points 𝑃 and 𝑄 on the circumference of a circle are a diameter if and only if they form a right triangle 𝑃𝑄𝑅 with any other point 𝑅 on the circumference. In this example, we proceed by trial and error: we pick one of the given chords and check to see if it forms a right triangle with one of the other points. If not, we pick another chord and try again.

Let us test the chord 𝐡𝐢 first. We will check to see if △𝐡𝐢𝐸 is a right triangle using the Pythagorean theorem, which states that △𝐡𝐢𝐸 is a right triangle if and only if 𝐡𝐸+𝐢𝐸=𝐡𝐢.

Let us calculate these lengths. We have 𝐡𝐸=(8βˆ’9)+(8βˆ’3)=26,𝐢𝐸=(8βˆ’4)+(8βˆ’8)=16, and 𝐡𝐢=(4βˆ’9)+(8βˆ’3)=50, but 𝐡𝐸+𝐢𝐸=42β‰ 50=𝐡𝐢, so △𝐡𝐢𝐸 is not a right triangle and 𝐡𝐢 is not a diameter.

Let us check 𝐢𝐷. We want to know if △𝐢𝐷𝐸 is a right triangle, so we check if 𝐷𝐸+𝐢𝐸=𝐢𝐷.

We have already calculated 𝐢𝐸=16, so we only need to work out the other two lengths. We have 𝐷𝐸=(8βˆ’8)+(2βˆ’8)=36 and 𝐢𝐷=(8βˆ’4)+(2βˆ’8)=52, and, indeed, 𝐷𝐸+𝐢𝐸=52=𝐢𝐷, confirming that △𝐢𝐷𝐸 is a right triangle and that 𝐢𝐷 is a diameter of the circle. Since we have identified the diameter and the question tells us that only one of the given chords is a diameter, there is no need to check 𝐷𝐹.

We will now see how to work out the equation of a circle, given some points on its circumference, by establishing that the chord connecting two of those points is a diameter of the circle.

Example 2: Showing That Three Points Form a Right Triangle Inscribed in a Semicircle and Finding the Equation of the Circle

The points 𝑃(βˆ’2,4), 𝑄(βˆ’6,12), and 𝑅(0,10) lie on a circle.

  1. What type of angle is βˆ π‘ƒπ‘…π‘„?
  2. Write down an equation for the circle in the form (π‘₯Β±π‘Ž)+(𝑦±𝑏)=𝑐, where π‘Ž, 𝑏, and 𝑐 are constants to be determined.

Answer

Part 1

We can find what type of angle βˆ π‘ƒπ‘…π‘„ is using the Pythagorean theorem, which states that βˆ π‘ƒπ‘…π‘„=90∘ if and only if 𝑃𝑅+𝑅𝑄=𝑃𝑄.

Calculating lengths, we find 𝑃𝑅=(0βˆ’(βˆ’2))+(10βˆ’4)=40,𝑅𝑄=(βˆ’6βˆ’0)+(12βˆ’10)=40, and 𝑃𝑄=(βˆ’6βˆ’(βˆ’2))+(12βˆ’4)=80.

So, indeed, 𝑃𝑅+𝑅𝑄=80=𝑃𝑄, which shows that βˆ π‘ƒπ‘…π‘„ is a right angle.

Part 2

Since βˆ π‘ƒπ‘…π‘„ is a right angle, we know that 𝑃𝑄 must be a diameter of the circle. This allows us to identify the center of the circle as the midpoint of 𝑃𝑄 and its radius as half of 𝑃𝑄.

We calculate the coordinates of the midpoint 𝑀 of 𝑃𝑄 using the formula 𝑀(π‘₯,𝑦)=ο€½π‘₯+π‘₯2,𝑦+𝑦2=ο€½βˆ’2+(βˆ’6)2,4+122=(βˆ’4,8).

Now, we can write down the equation for the circle: (π‘₯βˆ’π‘₯)+(π‘¦βˆ’π‘¦)=π‘Ÿ(π‘₯+4)+(π‘¦βˆ’8)=π‘Ÿ, where π‘Ÿ is the length of the radius, which is, of course, half the length of the diameter 𝑃𝑄. We have already calculated 𝑃𝑄=80, so π‘Ÿ=𝑃𝑄2=√802 and π‘Ÿ=ο€Ώβˆš802=804=20, and the equation of the circle is (π‘₯+4)+(π‘¦βˆ’8)=20.

In the previous example, we saw how to find the center of circle, given some points on its circumference, when two of those points lie on a diameter. We will now use a different technique to find a circle’s center, given points on its circumference, in the situation where no two of those points lie on a diameter.

As discussed in the introduction, if we have three noncolinear points 𝐴, 𝐡, and 𝐢, then the point of intersection 𝑃 of any two of the perpendicular bisectors of the sides 𝐴𝐡, 𝐡𝐢, and 𝐴𝐢 is equidistant from all three vertices. The point 𝑃 is therefore the circumcenter of 𝐴, 𝐡, and 𝐢, that is, the center of the circumcircle, the unique circle passing through 𝐴, 𝐡, and 𝐢.

If we are given that the points 𝐴, 𝐡, and 𝐢 lie on a circle, then the circumcenter 𝑃 must be the center of that circle. We can sum this up in the following property.

Property: Perpendicular Bisectors of Chords

The perpendicular bisectors of any two distinct chords of a circle intersect at the circle’s center.

Example 3: Finding the Perpendicular Bisector of a Chord of the Circle and Using It to Find the Center of the Circle

The points 𝐴(βˆ’3,βˆ’3), 𝐡(βˆ’5,βˆ’7), and 𝐢(βˆ’9,βˆ’3) lie on the circumference of a circle. The equation of the perpendicular bisector of 𝐡𝐢 is 𝑦=π‘₯+2. Find the equation of the perpendicular bisector of 𝐴𝐡 and the coordinates of the center of the circle.

Answer

To find the equation of the perpendicular bisector of a chord, we need two pieces of information: the slope of the chord and the coordinates of its midpoint. The slope of the chord 𝐴𝐡 is given by slopeofchangeinchangein𝐴𝐡=𝑦π‘₯=βˆ’7βˆ’(βˆ’3)βˆ’5βˆ’(βˆ’3)=βˆ’4βˆ’2=2.

The slope of the perpendicular bisector is then the negative reciprocal of the slope of the chord, so, in this case, it is βˆ’12. We can now write down the following equation for the perpendicular bisector 𝑦=βˆ’12π‘₯+𝑏, and we can calculate the value of the constant 𝑏 using the coordinates of the midpoint of the chord, which are given by 𝑀(π‘₯,𝑦)=ο€Όπ‘₯+π‘₯2,𝑦+𝑦2=ο€½βˆ’3+(βˆ’5)2,βˆ’3+(βˆ’7)2=(βˆ’4,βˆ’5).

Substituting into 𝑦=βˆ’12π‘₯+𝑏, we have 𝑦=βˆ’12π‘₯+π‘βˆ’5=βˆ’12Γ—(βˆ’4)+𝑏𝑏=βˆ’5βˆ’2=βˆ’7, giving us the equation 𝑦=βˆ’12π‘₯βˆ’7 for the perpendicular bisector of 𝐴𝐡.

The center of the circle lies at the intersection of the two perpendicular bisectors 𝑦=π‘₯+2 and 𝑦=βˆ’12π‘₯βˆ’7. Solving for π‘₯, we get π‘₯+2=βˆ’12π‘₯βˆ’7π‘₯+12π‘₯=βˆ’7βˆ’232π‘₯=βˆ’9π‘₯=βˆ’6, and substituting π‘₯=βˆ’6 back into 𝑦=π‘₯+2, we have 𝑦=βˆ’6+2=βˆ’4.

Thus, the coordinates of the center of the circle are (βˆ’6,βˆ’4).

In fact, once we have used the perpendicular bisectors of chords of a circle to find its center, we simply need to find the square of its radius to write an equation for the circle. This is what we are going to do in the next example.

Example 4: Using the Bisectors of Two Chords of a Circle to Find the Equation of the Circle

The points π‘ˆ(4,βˆ’2), 𝑉(βˆ’2,4), and π‘Š(βˆ’8,βˆ’6) lie on a circle. Given that the equations of the perpendicular bisectors of π‘ˆπ‘‰ and π‘ˆπ‘Š are 𝑦=π‘₯ and 𝑦=βˆ’3π‘₯βˆ’10, write down an equation for the circle in the form (π‘₯Β±π‘Ž)+(𝑦±𝑏)=𝑐, where π‘Ž, 𝑏, and 𝑐 are constants to be determined.

Answer

Since we are given the equations 𝑦=π‘₯ and 𝑦=βˆ’3π‘₯βˆ’10 of the perpendicular bisectors of two distinct chords of the circle, we can immediately find the center 𝐢 of the circle at their intersection. Substituting 𝑦=π‘₯ into 𝑦=βˆ’3π‘₯βˆ’10, we have π‘₯=βˆ’3π‘₯βˆ’104π‘₯=βˆ’10π‘₯=βˆ’52, and since 𝑦=π‘₯, the coordinates of the center are πΆο€Όβˆ’52,βˆ’52. The only other thing we need to write down an equation for the circle is the square of the length of its radius, π‘ŸοŠ¨. This is simply the square of the distance from the center to any point on the circumference, say, π‘ˆ(4,βˆ’2). This square distance is given by πΆπ‘ˆ=ο€Ό4βˆ’ο€Όβˆ’52+ο€Όβˆ’2βˆ’ο€Όβˆ’52=ο€Ό132+ο€Ό12=1694+14=1704=852.

We can therefore write down the following equation for the circle: ο€Όπ‘₯+52+𝑦+52=852.

We are now in a position to calculate an equation of a circle given any collection of at least three distinct points on its circumference using the perpendicular bisectors of any pair of chords connecting them.

Example 5: Finding the Equation of the Circle given Four Points on the Circumference

The points 𝐴(3,βˆ’7), 𝐡(βˆ’12,βˆ’4), 𝐢(βˆ’8,4), and 𝐷(4,βˆ’4) lie on a circle. By finding the perpendicular bisectors of 𝐴𝐡 and 𝐢𝐷, write down an equation for the circle in the form (π‘₯Β±π‘Ž)+(𝑦±𝑏)=𝑐, where π‘Ž, 𝑏, and 𝑐 are constants to be determined.

Answer

To write down equations of the perpendicular bisectors of 𝐴𝐡 and 𝐢𝐷, we need to work out the slopes and midpoints of these chords. The slope of the chord 𝐴𝐡 is given by slopeofchangeinchangein𝐴𝐡=𝑦π‘₯=βˆ’4βˆ’(βˆ’7)βˆ’12βˆ’3=βˆ’315=βˆ’15, and the slope of 𝐢𝐷 is given by slopeof𝐢𝐷=βˆ’4βˆ’44βˆ’(βˆ’8)=βˆ’812=βˆ’23.

We have two equations for bisectors, 𝑦=5π‘₯+𝑏𝑦=32π‘₯+𝑏,and where the slopes are the negative reciprocals of the corresponding slopes of the chords and the constants π‘οŒ οŒ‘ and π‘οŒ’οŒ£ are to be determined using, for each bisector, a point lying on it: its midpoint.

The midpoint of 𝐴𝐡 is given by 𝑀=ο€Όπ‘₯+π‘₯2,𝑦+𝑦2=ο€Ό3βˆ’122,βˆ’7βˆ’42=ο€Όβˆ’92,βˆ’112, and the midpoint of 𝐢𝐷 is given by 𝑀=ο€Όπ‘₯+π‘₯2,𝑦+𝑦2=ο€Όβˆ’8+42,4βˆ’42=(βˆ’2,0).

Substituting the coordinates of π‘€οŒ οŒ‘ into 𝑦=5π‘₯+π‘οŒ οŒ‘, we find βˆ’112=5Γ—βˆ’92+𝑏𝑏=452βˆ’112=342=17.

A similar calculation yields 𝑏=3. So, the two perpendicular bisectors are given by 𝑦=5π‘₯+17𝑦=32π‘₯+3.and

Since we are looking for the intersection of these two lines (that is, the center of the circle), we equate the 𝑦-coordinates and solve for π‘₯: 5π‘₯+17=32π‘₯+35π‘₯βˆ’32π‘₯=3βˆ’1772π‘₯=βˆ’14π‘₯=βˆ’4, and then, we substitute π‘₯=βˆ’4 back into 𝑦=5π‘₯+17 for 𝑦=5Γ—(βˆ’4)+17=βˆ’20+17=βˆ’3.

Hence, the coordinates of the center of the circle are (βˆ’4,βˆ’3) and we have the following equation for the circle: (π‘₯+4)+(𝑦+3)=π‘Ÿ, where π‘ŸοŠ¨ is the square of the radius of the circle. We can calculate π‘ŸοŠ¨ as the square of the distance from the center (βˆ’4,βˆ’3) to any point on the circumference, say, 𝐴(3,βˆ’7), by π‘Ÿ=(3βˆ’(βˆ’4))+(βˆ’7βˆ’(βˆ’3))=49+16=65.

So, the equation of the circle is given by (π‘₯+4)+(𝑦+3)=65.

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

Key Points

  • There is a unique circle passing through any three noncolinear points 𝐴, 𝐡, and 𝐢 in the plane. It is called the circumcircle of △𝐴𝐡𝐢.
  • The center of the circumcircle is called the circumcenter. It is equidistant from the points 𝐴, 𝐡, and 𝐢 and lies at the intersection point of the perpendicular bisectors of the sides 𝐴𝐡, 𝐡𝐢, and 𝐴𝐢.
  • Three points in the plane form the vertices of a right triangle if and only if the hypotenuse is the diameter of the circumcircle through the vertices. We can use this fact to calculate the equation of the circumcircle.
  • The perpendicular bisectors of any two distinct chords of a circle intersect in the circle’s center. We can use this fact to calculate the equation of a circle given any collection of at least three distinct points on its circumference.

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