## Engage NY Eureka Math Algebra 2 Module 1 Lesson 32 Answer Key

### Eureka Math Algebra 2 Module 1 Lesson 32 Example Answer Key

Rewrite x^{2} + y^{2} – 4x + 2y = – 1 by completing the square in both x and y. Describe the circle represented by this equation.

Answer:

Rearranging terms gives x^{2} – 4x + y^{2} + 2y = -1.

Then, completing the square in both x and y, we have

(x^{2} – 4x + 4) + (y^{2} + 2y + 1) = – 1 + 4 + 1

(x – 2)^{2} + (y + 1)^{2} = 4.

This is the equation of a circle with center (2, – 1) and radius 2.

Using your graphing tool, graph the circle.

Answer:

See the graph to the right.

In contrast, consider the following equation: x^{2} + y^{2} – 2x – 8y = – 19.

Answer:

Rearranging terms gives x^{2} – 2x + y^{2} – 8y = – 19.

Then, completing the square in both x and y, we have

(x^{2} – 2x + 1) + (y^{2} – 8y + 16) = – 19 + 1 + 16

(x – 1)^{2} + (y – 4)^{2} = – 2,

which is not a circle because then the radius would be âˆš2.

What happens when you use your graphing tool with this equation?

Answer:

The tool cannot draw the graph. There are no points in the plane that satisfy this equation, so the graph is empty.

Example 2.

Find the distance between the centers of the two circles with equations below, and use that distance to determine in how many points these circles intersect.

x^{2} + y^{2} = 5

(x – 2)^{2} + (y – 1)^{2} = 3

Answer:

The first circle has center (0, 0), and the second circle has center (2,1). Using the distance formula, the distance between the centers of these circles is

d = \(\) = âˆš5

Since the distance between the centers is between the sum and the difference of the two radii, that is,

âˆš5 – âˆš3 < âˆš5 < âˆš5 + âˆš3, we know that the circles must intersect in two distinct points.

Example 3.

Point A(3, 2) is on a circle whose center is C(- 2, 3). What is the radius of the circle?

Answer:

The distance from A to c is given by \(\sqrt{(3+2)^{2}+(2-3)^{2}}\) = âˆš26, which is the length of the radius.

What is the equation of the circle? Graph it.

Answer:

Given the center and the radius, we can write the equation of the circle as

(x + 2)^{2} + (y – 3)^{2} = 26.

The graph is shown at the right.

Use the fact that the tangent at A(3, 2) is perpendicular to the radius at that point to find the equation of the tangent line. Then graph it.

Answer:

The slope of the tangent line is the opposite reciprocal of the slope of . The slope of is \(\frac{3-2}{-2-3}\) = –\(\frac{1}{5}\), so the slope of the tangent line is 5. Using the point-slope form of the equation of a line with slope 5 and passing through point (3, 2) gives

y – 2 = 5(x – 3)

y = 5x – 13.

The equation of the tangent line is, therefore, y = 5x – 13.

Find the coordinates of point B, the second intersection of the and the circle.

Answer:

The system (x + 2)^{2} + (y – 3)^{2} = 26 and 5y = – x + 13 can be solved by substituting x = 13 – 5y into the equation of

the circle, which yields (13 – 5y + 2)^{2} + (y – 3)^{2} = 26. This gives 26(y – 2) (y – 4) = 0. Thus, the y-coordinate is either 2 or 4. If y = 2, then x = 13 – 5 âˆ™ 2 = 3, and if y = 4, then x = 13 – 5 âˆ™ 4 = – 7. Since A has coordinates (3, 2), it follows that B has coordinates (- 7, 4).

What is the equation of the tangent to the circle at (- 7, 4)? Graph it as a check.

Answer:

Using the point-slope form of a line with slope 5 and point (- 7, 4):

y – 4 = 5(x + 7)

y = 5x + 39.

The equation of the tangent line is, therefore, y = 5x + 39.

The graph is shown to the right.

The lines y = 5x + b are parallel to the tangent lines to the circle at points A and B. How is the y-intercept b for these lines related to the number of times each line intersects the circle?

Answer:

When b = – 13 or b = 39, the line is tangent to the circle, intersecting in one point.

When – 13 < b < 39, the line intersects the circle in two points.

When b < – 13 or b > 39, the line and circle do not intersect.

### Eureka Math Algebra 2 Module 1 Lesson 32 Opening Exercise Answer Key

Given the line y = 2x, is there a point on the line at a distance 3 from (1, 3)? Explain how you know.

Answer:

Yes, there are two such points. They are the intersection of the line y = 2x and the circle

(x – 1)^{2} + (y – 3)^{2} = 9. (The intersection points are roughly (0.07, 0. 15) and (2. 73, 5.45).)

Draw a graph showing where the point is.

Answer:

There are actually two such points. See the graph below.

### Eureka Math Algebra 2 Module 1 Lesson 32 Exercise Answer Key

Exercise 1.

Solve the system (x – 1)^{2} + (y – 2)^{2} = 2^{2} and y = 2x + 2.

Answer:

Substituting 2x + 2 for y in the quadratic equation allows us to find the x-coordinates.

(x – 1)^{2} +((2x+ 2) – 2)^{2} = 4

(x^{2} – 2x + 1) + 4x^{2} = 4

5x^{2} – 2x – 3 = 0

(x – 1) (5x + 3) = 0

So, x = 1 or x = –\(\frac{3}{5}\), and the intersection points are \(\left(-\frac{3}{5}, \frac{4}{5}\right)\), and (1, 4).

What are the coordinates of the center of the circle?

Answer:

(1, 2)

What can you say about the distance from the intersection points to the center of the circle?

Answer:

Because they are points on the circle and the radius of the circle is 2, the intersection points are 2 units away from the center. This can be verified by the distance formula.

Using your graphing tool, graph the line and the circle.

Answer:

See the graph below.

Exercise 2.

Consider a circle with radius 5 and another circle with radius 3. Let d represent the distance between the two centers. We want to know how many intersections there are of these two circles for different values of d. Draw figures for each case.

a. What happens if d = 8?

Answer:

If the distance is 8, then the circles touch at only one point. We say that the circles are externally tangent.

b. What happens if d = 10?

Answer:

If the distance is 10, the circles do not intersect, and neither circle is inside the other.

c. What happens if d = 1?

Answer:

If the distance is 1, the circles do not intersect, but one circle lies inside the other.

d. What happens if d = 2?

Answer:

If the distance is 2, the circles touch at only one point, with one circle inside the other. We say that the circles are internally tangent.

e. For which values of d do the circles intersect in exactly one point? Generalize this result to circles of any radius.

Answer:

If d = 8 or d = 2, the circles are tangent. In general, if d is either the sum or the difference of the radii, then the circles are tangent.

f. For which values of d do the circles intersect in two points? Generalize this result to circles of any radius.

Answer:

If 2 < d < 8, the circles intersect in two points. In general, if d is between the sum and the difference of the radii, then the circles intersect in two points.

g. For which values of d do the circles not intersect? Generalize this result to circles of any radius.

Answer:

The circles do not intersect if d < 2 or d > 8. In general, if d is smaller than the difference of the radii or larger than the sum of the radii, then the circles do not intersect.

Exercise 3.

Use the distance formula to show algebraically and graphically that the following two circles do not intersect.

(x – 1)^{2} + (y + 2)^{2} = 1

(x + 5)^{2} + (y – 4)^{2} = 4

The centers of the two circles are (1, – 2) and (- 5, 4), and the radii are 1 and 2. The distance between the two centers is \(\sqrt{6^{2}+6^{2}}=6 \sqrt{2}\) which is greater than 1 + 2 = 3. The graph below also shows that the circles do not intersect.

### Eureka Math Algebra 2 Module 1 Lesson 32 Problem Set Answer Key

Question 1.

Use the distance formula to find the distance between the points (- 1, – 13) and (3, – 9).

Answer:

Using the formula with (- 1, – 13) and (3, – 9),

Therefore, the distance is 4âˆš2.

Question 2.

Use the distance formula to find the length of the longer side of the rectangle whose vertices are (1, 1), (3, 1), (3, 7), and (1, 7).

Answer:

Using the formula with (1, 1) and (1, 7),

d = \(\sqrt{(1-1)^{2}+(7-1)^{2}}\)

d = \(\sqrt{(0)^{2}+(6)^{2}}\) = âˆš36

Therefore, the length of the longer side is 6.

Question 3.

Use the distance formula to find the length of the diagonal of the square whose vertices are (0, 0), (0, 5), (5, 5), and (5, 0).

Answer:

Using the formula with (0, 0) and (55),

d = \(\sqrt{(5-0)^{2}+(5-0)^{2}}\)

d = \(\sqrt{(5-0)^{2}+(5-0)^{2}}\) = \(\sqrt{25+25}\) = 5âˆš2.

Therefore, the length of the diagonal is 5âˆš2.

Write an equation for the circles in Exercises 4 – 6 in the form (x – h)^{2} + (y – k)^{2} = r^{2}, where the center is (h, k) and the radius is r units. Then write the equation in the standard form x^{2} + ax + y^{2} + by + c = 0, and construct the graph of the equation.

Question 4.

A circle with center (4, – 1) and radius 6 units.

Answer:

(x – 4)^{2} + (y + 1)^{2} = 36; standard form: x^{2} – 8x + y^{2} + 2y – 19 = 0

The graph is shown to the right.

Question 5.

A circle with center (-3, 5) tangent to the x-axis.

Answer:

(x + 3)^{2} + (y – 5)^{2} = 25; standard form: x^{2} + 6x + y^{2} – 10y + 9 = 0

The graph is shown to the right.

Question 6.

A circle in the third quadrant, radius 1 unit, tangent to both axes.

Answer:

(x + 1)^{2} + (y + 1)^{2} = 1; standard form: x^{2} + 2x + y^{2} + 2y + 1 = 0

The graph is shown to the right.

Question 7.

By finding the radius of each circle and the distance between their centers, show that the circles x^{2} + y^{2} = 4 and x^{2} – 4x + y^{2} – 4y + 4 = 0 intersect. Illustrate graphically.

Answer:

The second circle is (x – 2)^{2} + (y – 2)^{2} = 4. Each radius is 2, and the

centers are at (0, 0) and (2, 2). The distance between the centers is 2âˆš2, which is less than 4, the sum of the radii.

The graph of the two circles is to the right.

Question 8.

Find the points of intersection of the circles x^{2} + y^{2}– 15 = 0 and x^{2} – 4x + y^{2} + 2y – 5 = 0. Check by graphing the equations. Write the equations as

Answer:

x^{2} + y^{2} = 15

x^{2} + y^{2} – 4x + 2y = 5.

Subtracting the second equation from the first

4x – 2y = 10,

which is equivalent to

2x – y = 5.

Solving the system x^{2} + y^{2} = 15 and y = 2x – 5 yields

(2 + âˆš2, – 1 + 2âˆš2) and (2 – âˆš2, – 1 – 2âˆš2). The graph is to the right.

Question 9.

Solve the system y = x^{2} – 2 and x^{2} + y^{2} = 4. Illustrate graphically.

Answer:

Substitute x^{2} = y + 2 into the second equation:

y + 2 + y^{2} = 4

y^{2} + y – 2 = 0

(y – 1) (y + 2) = 0

so y = – 2 or y = 1.

If y = – 2, then x^{2} = y + 2 = 0, and thus x = 0.

If y = 1, then x^{2} = y + 2 = 3, so x = âˆš3 or x = – âˆš3

Thus, there are three solutions (0, – 2), (âˆš3, 1), and (-âˆš3, 1). The graph is to the right.

Question 10.

Solve the system y = 2x – 13 and y = x^{2} – 6x + 3. Illustrate graphically.

Answer:

Substitute 2x – 13 for y in the second equation: 2x – 13 = x^{2} – 6x + 3.

Rewrite the equation in standard form: x^{2} – 8x + 16 = 0.

Solve for x: (x – 4)(x -4) = 0.

The root is repeated, so there is only one solution x = 4.

The corresponding y-value is y = -5, and there is only one solution, (4,-5).

As shown to the right, the line is tangent to the parabola.

### Eureka Math Algebra 2 Module 1 Lesson 32 Exit Ticket Answer Key

Question 1.

Find the intersection of the two circles

x^{2} + y^{2} – 2x + 4y – 11 = 0

and

x^{2} + y^{2} + 4x + 2y – 9 = 0.

Answer:

Subtract the second equation from the first: – 6x + 2y – 2 = 0.

Solve the equation for y: y = 3x + 1.

Substitute in the first equation: x^{2} + (3x + 1)^{2} – 2x + 4(3x + 1) – 11 = 0.

Remove parentheses, and combine like terms: 5x^{2} + 8x – 3 = 0.

Substitute in the quadratic equation to find two values: x = \(\frac{-4}{5}-\frac{\sqrt{31}}{5}\), and x = \(\frac{-4}{5}+\frac{\sqrt{31}}{5}\)

The corresponding y-values are the following: y = \(\frac{-7}{5}-\frac{3 \sqrt{31}}{5}\), and y = \(\frac{-7}{5}+\frac{3 \sqrt{31}}{5}\).

Question 2.

The equations of the two circles in Question 1 can also be written as follows:

(x – 1)^{2} + (y + 2)^{2} = 16

and

(x + 2)^{2} + (y + 1)^{2} = 14.

Graph the circles and the line joining their points of intersection.

Answer:

See the graph to the right.

Question 3.

Find the distance between the centers of the circles in Questions 1 and 2.

Answer:

The center of the first circle is (1, – 2), and the center of the second circle is (- 2, – 1). We then have

d = \(\sqrt{(-2-1)^{2}+(-1+2)^{2}}\) = \(\sqrt{9+1}\) = âˆš10.