Transformations - Circles: Constructions and Equations - High School Geometry Unlocked (2016)

High School Geometry Unlocked (2016)

Chapter 8. Circles: Constructions and Equations

Lesson 8.3. Transformations

In this section, you will apply what you’ve learned about circles to problems, theorems, and proofs.

In the figure above, the circle is in the standard coordinate plane, with center (2, 7) and radius 13.

For each of the coordinates below, determine whether the point is inside the circle, outside the circle, or on the circle.

1. (−3, 19)

3. (16, 14)

5. (−10, 12)

2. (6, −5)

4. (−14, 2)

6. (5, 14)

Anytime you’re asked to prove whether or not a point is on a circle, the question you’re really being asked is distance. Your task is to determine whether or not the distance for each pair of coordinates is equal to the radius of the circle.

We know that the radius of the circle is 13. That means that if a coordinate is on the circle, then it is exactly 13 units from the center. What does it mean if the coordinate is less than 13 units from the center? More than 13?

Use the distance formula to find the distance between each point and the center of the circle (2, 7).

We’ll do #1 step by step. Try the rest on your own, and then check your answers against ours.

1.

The distance is 13 (equal to the radius), so this point is on the circle.

2.

The distance is ≈ 12.65 (less than the radius), so this point is inside the circle.

3.

The distance is ≈ 15.65 (greater than the radius), so this point is outside the circle.

4.

The distance is ≈ 16.76 (greater than the radius), so this point is outside the circle.

5.

The distance is 13 (equal to the radius), so this point is on the circle.

6.

The distance is ≈ 7.62 (less than the radius), so this point is inside the circle.

Here is how you may see the equation of a circle on the SAT.

The standard form of the equation of a circle is (xh)2 + (yk)2 = r2, where the center of the circle is at point (h, k) and the radius of the circle is r. What is the standard form of the equation of the circle defined by the equation x2 + y2 − 6x + 8y = 0?

A.(x − 6)2 + (y + 8)2 = 0

B.(x + 3)2 + (y + 4)2 = 25

C.(x − 3)2 + (y + 4)2 = 25

D.(x − 3)2 + (y + 4)2 = 5

In the figure above, the circle has the equation x2 + y2 = 52. Line p passes through the center of the circle, and it has the equation y = x.

Line q (not shown) has the equation y = , and it passes through the point (3, 4). Is line q tangent to the circle?

Recall the definition of a tangent line: a line that intersects the circle at a single point, and is perpendicular to the radius at the point of tangency. In order to determine whether the line is tangent, we’ll need to prove both of those things.

Perhaps an easy place to start is to compare the equations of the two lines. Since line p passes through the center of the circle, we know that a radius exists on that line. Therefore, it should be perpendicular to line q, if line q is tangent to the circle.

Recall that two lines are perpendicular if they have opposite reciprocal slopes.

The slopes of lines p and q are and , respectively. Are those opposite reciprocals? Yes! This is proof that the lines could be tangent.

We will also need to know if the lines intersect at the point of tangency. We could solve the two linear equations as a system. However, we were given a point for line q (3, 4). Why not try that one? Let’s see if it satisfies the equation for line p.

y = x

4 = × 3

Substitute (3, 4) for (x, y).

4 = 4

True!

This means that the two lines intersect at point (3, 4).

Now, try the point (3, 4) in the equation for the circle. If the point lies on the circle, that means that it is an intersection point for both lines as well as the circle.

x2 + y2 = 52

32 + 42 = 52

Substitute (3, 4) for (x, y).

9 + 16 = 25

True!

To recap, we now know that the point (3, 4) is the intersection of both lines and the circle. We also know that lines p and q are perpendicular to each other. Therefore, we have proved that line q is tangent to the circle.

Start by rewriting the equation with the x terms and y terms listed together:

x2 − 6x + y2 + 8y = 0

Next, to get the equation into standard form, you want to complete the square. Start with the x terms. Take half of the coefficient on the x (−6), square it, and add that to both sides. Half of −6 is 3, and 32 is 9, so add 9 to both sides of the equation:

x2 − 6x + 9 + y2 + 8y = 9

Now x2 − 6x + 9 is a perfect square; it factors into (x − 3)2, so you can rewrite the equation:

(x − 3)2 + y2 + 8y = 9

You can do the same to the y terms. Half of 8 is 4, and 42 = 16, so add 16 to both sides of the equation:

(x − 3)2 + y2 + 8y + 16 = 9 + 16

Now y2 + 8y + 16 is a perfect square; it factors into (x + 4)2, so you can rewrite the equation again:

(x − 3)2 + (y + 4)2 = 25

That matches (D). Note you do not need to take the square root of 25; that would change the value of the equation. Instead, the radius is left squared. The correct answer is (C).

Given:

AB is a diameter of the circle.

C is a point on the circle.

O is the center of the circle.

Prove:

ACB is 90°

One way to prove that ∠CAB is 90° is to use the fact that all radii are equal. Since C is a point on the circle, we know that OC is a radius. That means that OC is congruent to OA and OB.

Because the two radii are equal, we know that the two smaller triangles are both isosceles. That means that each one has a pair of congruent base angles.

We also know that the angles in triangle ABC must have a sum of 180°. Therefore, we can set up an equation:

x + y + x + y = 180°

2x + 2y = 180°

Simplify.

2 (x + y) = 180°

x + y =

x + y = 90°

We have proved that the angle ACB, which is equal to x + y, must be equal to 90°.

Given:

O is the center of the circle.

AC is tangent to the circle at point A.

BC is tangent to the circle at point B.

Prove:

ACBC

One way to prove that ACBC is by using congruent triangles. If we split the quadrilateral down the middle, we’d have two right triangles, as shown:

We know that OAOB, because they are both radii. Also, we know of course that OCOC, by the Reflexive Property.

Therefore, we have two right triangles with two pairs of congruent, corresponding sides. The triangles must be congruent. You can refer to the Hypotenuse-Leg Theorem, which states that if two right triangles have a hypotenuse and leg pair that are congruent, then the two triangles are congruent.

Finally, if the two triangles are congruent, then we know that the corresponding sides ACBC must be congruent.

DRILL

CHAPTER 8 PRACTICE QUESTIONS

Directions: Complete the following problems as specified by each question. For extra practice after answering each question, try using an alternative method to solve the problem or check your work. Larger, printable versions of images are available in your online student tools.

Click here to download a PDF of Chapter 8 Practice Questions.

1.Using a compass and a straightedge, construct an equilateral triangle with side AB.

2.Using a compass and a straightedge, construct the circumcircle of right triangle ABC below.

3.Using a compass and a straightedge, construct the incircle of the regular hexagon shown below.

4.Is the point (−3, −2) inside, outside, or on the circle x2 + y2 − 6x + y2 + 8y = 39?

5.Graph the circle represented by the equation x2 + y2 + 4x − 2y = 20.

6.In the figure below, XY and XZ are tangent to the circle with center O at Y and Z, respectively, and ∠YXZ is a right angle.

Is ΔXYZ isosceles? Why?

7.In the figure below, diameters AC and BD intersect at point O, and O is the center of the circle.

Is ABCD? Why?

8.In the figure below, AB is tangent to circles P and Q at points A and B, respectively, and DC is tangent to circles P and Q at points D and C, respectively. AD is a diameter of circle P and BC is a diameter of circle Q.

Are circles P and Q are congruent? Why?

SOLUTIONS TO CHAPTER 8 PRACTICE QUESTIONS

1.An equilateral triangle has three congruent sides. To form an equilateral triangle, find a point (call it C for reference) such that ABBCCA. To make sure that ABCA, construct a circle with center A and radius AB.

Any other radius of this circle is congruent to AB. To make sure that ABBC, construct a circle with center A and radius AB.

Any other radius of this circle is congruent to AB. There are two points that are on both circles. Use the upper point. Call that C. AB is a radius of both circle A and circle B, so it is congruent to both AC and CB, so triangle ABC is equilateral.

The same would be true if you used the bottom intersection point as the third vertex.

2.To draw a circumcircle of a triangle, first find the circumcenter. In the case of a right triangle, there is a shortcut: The circumcenter is always the midpoint of the hypotenuse. To find the midpoint of a segment, construct the perpendicular bisector. To do this use a compass to construct a circle with needle point on A and drawing point on C. Now reverse this. Construct a circle with needle point on C and drawing point on A. The circles have two points of intersection. Use a straight edge to draw a line to connect the points of intersection.

Find the point that the bisector intersects the hypotenuse. That is the midpoint. Place the needle point of the compass here and the drawing point on any of the three vertices of the triangle. Construct this circle.

3.A polygon has an incircle if opposite sides are congruent. Since the hexagon is a regular hexagon, all sides (and thus opposite sides) are congruent, so this figure does have an incircle. To find the incircle, find the incenter, which is the intersection of the figure’s angle bisectors. Pick two points. As an example, use the top two vertices on the hexagon. Start with the top left. Construct a circle with the compass needle on the top left vertex through the two adjacent sides. Construct congruent circles with the compass needle on the points of intersection. Use a straight edge to construct a line through the vertex and the point of intersection between the second two circles.

Now do the same for the top right vertex.

The point of intersection of these two lines is the incenter. An incircle is tangent to any of the sides of the hexagon. To find the point of tangency, construct a perpendicular line from the incenter to any side. Use the compass to draw a circle through two points on one side of the circle. For example, use the bottom side. Now, construct two circles, one with the drawing point at the incenter and the needle point at one intersection point and another with the drawing point at the incenter and the needle point at the other intersection point. Draw a line through the incenter and the other intersection point on these two circles. It crosses the side of the hexagon at the tangent point.

Finally, with the needle point at the incenter and the drawing point at the tangent point, construct an incircle of the hexagon.

4.Inside.

Consider the definition of a circle: the set of points that is a given distance from a given point. That distance is called the radius, and that point is called the center. If a point is a distance equal to the radius from the center, the point is on the circle. If a point is a distance less than the radius from the center, the point is inside the circle. If a point is a distance greater than the radius from the center, the point is outside the circle. Determine the center and the radius of this circle. In the standard form of the equation, (xh)2 + (yk)2 = r2, the center is (h, k) and the radius is r. The equation is given in expanded form, so put it into standard form using the complete the square method. Get the x terms together and the y terms together to get (x2 − 6x) + (y2 + 8y) = 39. Add the square of half the coefficient on x to both sides. The coefficient on x is −6, half of −6 is −3, and the square of −3 is 9, so add 9 to both sides to get (x2 − 6x + 9) + (y2 + 8y) = 48. Add the square of half the coefficient on y to both sides. The coefficient on y is 8, half of 8 is 4, and the square of 4 is 16, so add 16 to both sides to get (x2 − 6x + 9) + (y2 + 8y + 16) = 64. Factor the expression in each set of parentheses to get (x − 3)2 + (y + 4)2 = 82. Therefore, the center is (3, −4) and the radius is 8.

Next, determine how far the point (−3, −2) is from the point (3, −4). Use the distance formula: . Since 36 < 40 < 49, 6 < < 7. Therefore, the distance between (−3, −2) and the center is less than the length of the radius, so (−3, −2) is inside the circle.

5.To graph a circle, get its center and radius. To do this, get the equation of the circle into standard form: (xh)2 + (yk)2 = r2 with center (h, k) and radius r. The equation is given in expanded form, so complete the square. First, group the x terms and the yterms: (x2 + 4x) + (y2 − 2y) = 20. Now, add the square of half the coefficient on x to both sides. The coefficient on x is 4, half is 2, and the square of 2 is 4, so add 4 to both sides to get (x2 + 4x + 4) + (y2 − 2y) = 24. Now, add the square of half the coefficient on yto both sides. The coefficient on y is −2, half is −1, and the square of −1 is 1, so add 1 to both sides to get (x2 + 4x + 4) + (y2 − 2y + 1) = 25. Factor the two expressions in parentheses to get (x + 2)2 + (y − 1)2 = 52. Therefore, the center of the circle is (−2, 1) and the radius is 5. Draw the center of the circle at (−2, 1) and draw points on the circle that are 5 away from the center.

Take advantage of the fact that 5 is the hypotenuse of a 3:4:5 right triangle. One point on the circle is 3 units to the right of and 4 units above the center: (−2 + 3, 1 + 4) = (1, 5). And so on: You can also find the points (1, −3), (−5, −3), (−5, 5), (2, 4), (2, −2), (−6, −2), and (−6, 4).

Also, plot the points 5 units above, 5 units below, 5 units to the left, and 5 units to the right of center: (−2, 6), (−2, −4), (−7, 1), and (3, 1). Plot these points on the graph and draw a smooth curve to connect them.

6.Yes.

The goal is to prove that ΔXYZ is isosceles. Notice that ΔXYZ is a right triangle. The hypotenuse of a right triangle cannot be congruent to another side in the triangle. Therefore, to show that it is isosceles, prove that YXXZ. The question says that YX is a tangent to the circle at Y, so draw a radius to Y and mark it as a right angle. The question also says that XZ is a tangent to the circle at Z, so draw a radius to Z and mark it as a right angle. This forms quadrilateral OXYZ, which has three right angles. Since the sum of the angles in a quadrilateral is 360, the fourth angle must also be a right angle, making the figure a rectangle. Since OZ and OY are both radii, they are congruent. If adjacent sides of a rectangle are congruent, the rectangle is a square. If the rectangle is a square, then YXXZ, which, as described above, proves that ∠XYZ is a 45° angle.

7.Yes.

The goal is to prove that ABCD. Each of these segments is part of a triangle. When this is the case, the method will often be to prove that the two triangles are congruent. Try to find a way to use one of the congruence postulates/theorems: SSS, SAS, ASA, or AAS. Find congruent corresponding sides and angles. The center of the circle is O, so AO, BO, CO, and DO are all radii and thus congruent. Since AC and BD intersect at O, ∠AOB and∠DOC are vertical angles and thus are congruent. Therefore, AODO, BOCO, and ∠AOB ≅ ∠DOC. Thus, the SAS postulate can be used to prove that triangles AOB and DOC are congruent. Thus, ABCD.

8.Yes.

The information in the question says that AB is tangent to circles P and Q at points A and B, respectively, and that DC is tangent to circles P and Q at points D and C, respectively. Since tangents, by definition, are perpendicular to the radii at the point of intersection, draw the four radii and mark them as perpendicular. Notice that AP and PD form diameter AD and that BQ and QC form diameter BC. Notice further that these segments form quadrilateral ABCD and that this quadrilateral has four right angles. By definition, a quadrilateral with four right angles is a rectangle, so ABCD is a rectangle. The goal is to prove that the two circles are congruent. This can be done by showing that they have congruent radii. In this case, the diameters are opposite sides in a rectangle. By a property of a rectangle, opposite sides are congruent. Since the diameters are congruent, the radii must be congruent and so are the circles.

REFLECT

Congratulations on completing Chapter 8!

Here’s what we just covered.

Rate your confidence in your ability to:


•Construct the incenter, incircle, circumcenter, and circumcircle for triangles and polygons

•Construct the centroid and orthocenter for triangles

•Understand the standard equation of a circle and use the equation to solve for the center, radius, or coordinates on a circle

•Use the “completing the square” technique for a circle equation in expanded form

•Graph a circle from its equation

•Apply your understanding to general theorems and proofs

If you rated any of these topics lower than you’d like, consider reviewing the corresponding lesson, especially if you found yourself unable to correctly answer one of the related end-of-chapter questions.

Access your online student tools for a handy, printable list of Key Points for this chapter. These can be helpful for retaining what you’ve learned as you continue to explore these topics.