1. The Circle and the Parabola
Section 13.1
The Circle and the Parabola

2. Objectives Identify conic sections and some of their applications
Graph equations of circles written in standard form
Write the equation of a circle, given its center and radius
Convert the general form of the equation of a circle to standard form
Solve application problems involving circles
Convert the general form of the equation of a parabola to standard form to graph it

3. Objective 1: Identify Conic Sections and Some of Their Applications
The curves formed by the intersection of a plane with an infinite right-circular cone are called conic sections. Those curves have four basic shapes, called circles, parabolas, ellipses, and hyperbolas, as shown below.

4. Objective 1: Identify Conic Sections and Some of Their Applications

5. Objective 1: Identify Conic Sections and Some of Their Applications
Parabolas can be rotated to generate dish-shaped surfaces called paraboloids. Any light or sound placed at the focus of a paraboloid is reflected outward in parallel paths. This property makes parabolic surfaces ideal for flashlight and headlight reflectors. It also makes parabolic surfaces good antennas, because signals captured by such antennas are concentrated at the focus. Parabolic mirrors are capable of concentrating the rays of the sun at a single point, thereby generating tremendous heat. This property is used in the design of solar furnaces.
Any object thrown upward and outward travels in a parabolic path. An example of this is a stream of water flowing from a drinking fountain. In architecture, many arches are parabolic in shape, because this gives them strength. Cables that support suspension bridges hang in the shape of a parabola.

6. Objective 1: Identify Conic Sections and Some of Their Applications

7. Objective 2: Graph Equations of Circles Written in Standard Form
Definition of a Circle: A circle is the set of all points in a plane that are a fixed distance from a fixed point called its center. The fixed distance is called the radius of the circle.
The standard form of the equation of a circle with radius r and center at (h, k) is
(x – h)2 + (y – k)2 = r2

8. Find the center and the radius of each circle and then graph it:
EXAMPLE 1
Strategy We will compare each equation to the standard form of the equation of a circle,
(x – h)2 + (y – k)2 = r2 and identify h, k and r.
Why The center of the circle is the point with coordinates (h, k) and the radius of the circle is r.

9. Find the center and the radius of each circle and then graph it:
EXAMPLE 1
Solution
a. The color highlighting shows how to compare the given equation to the standard form to find h, k, and r.
The center of the circle is (h, k) = (4, 1) and the radius is 3.
To plot four points on the circle, we move up, down, left, and right 3 units from the center, as shown in figure (a). Then we draw a circle through the points to get the graph of
(x – 4)2 + (y – 1)2 = 9, as shown in figure (b).

10. Find the center and the radius of each circle and then graph it:
EXAMPLE 1
Solution
b. To find h and k, we will write x2 + y2 = 25 in the following way:
The center of the circle is at (0, 0) and the radius is 5.
To plot four points on the circle, we move up, down, left, and right 5 units from the center. Then we draw a circle through the points to get the graph of x2 + y2 = 25, as shown.

11. Find the center and the radius of each circle and then graph it:
EXAMPLE 1
Solution
c. To find h, we will write x + 3 as x – (–3).
Since r2 = 12 , we have
Since the radius can’t be negative, The center of the circle is at (–3, 0) and the radius is

12. Find the center and the radius of each circle and then graph it:
EXAMPLE 1
Solution
c. continued…
To plot four points on the circle, we move up, down, left, and right units from the center. We then draw a circle through the points to get the graph of (x + 3)2 + y2 = 12, as shown in the graph.

13. Objective 3: Write the Equation of a Circle, Given Its Center and Radius
Because a circle is determined by its center and radius, that information is all we need to know to write its equation.

14. EXAMPLE 2
Write the equation of the circle with radius 9 and center at (6, –5).
Strategy We substitute 9 for r, 6 for h, and –5 for k in the standard form of the equation of a circle,
(x – h)2 + (y – k)2 = r2.
Why When writing the standard form, the center is represented by the ordered pair (h, k) and the radius as r.

15. EXAMPLE 2
Write the equation of the circle with radius 9 and center at (6, –5).
Solution
This is the standard form.
If we express 92 as 81, we can write the equation as

16. Objective 4: Convert the General Form of the Equation of a Circle to Standard Form
In Example 2, the result was written in standard form:
(x – 6)2 + (y + 5)2 = 81. If we square (x – 6) and (y + 5), we obtain a different form for the equation of the circle.
This result is written in the general form of the equation of a circle.

17. Objective 4: Convert the General Form of the Equation of a Circle to Standard Form
Equation of the Circle: The general form of the equation of a circle is
x2 + y2 + Dx + Ey + F = 0
We can convert from the general form to the standard form of the equation of a circle by completing the square.

18. EXAMPLE 3
Write the equation x2 + y2 − 4x + 2y − 11 = 0
in standard form and graph it.
Strategy We will rearrange the terms to write the equation in the form x2 – 4x + y2 + 2y = 11 and complete the square on x and y.
Why Standard form contains the expressions (x – h)2 and (y – k)2. We can obtain a perfect-square trinomial that factors as (x – 2)2 by completing the square on x2 – 4x. We can complete the square on y2 + 2y to obtain an expression of the form (y + 1)2.

19. EXAMPLE 3 Solution Write the equation x2 + y2 − 4x + 2y − 11 = 0
in standard form and graph it.
Solution
To write the equation in standard form, we complete the square twice.
To complete the square on x2 – 4x, we note that and (–2)2 = 4. To complete the square on y2 + 2y we note that and 12 = 1. We add 4 and 1 to both sides of the equation.

20. EXAMPLE 3 Solution continued…
Write the equation x2 + y2 − 4x + 2y − 11 = 0
in standard form and graph it.
Solution continued…
The equation can also be written as (x – 2)2 + (y + 1)2 = 42. We can determine the circle’s center and radius by comparing this equation to the standard form of the equation of a circle, (x – h)2 + (y – k)2 = r2. We see that h = 2, k = –1, and r = 4. We can use the center, (h, k) = (2, –1) and the radius r = 4, to graph the circle as shown below.

21. Objective 5: Solve Application Problems Involving Circles
Radio Translators. The broadcast area of a television station is bounded by the circle
x2 + y2 = 3600, where x and y are measured in miles. A translator station picks up the signal and retransmits it from the center of a circular area bounded by
(x + 30)2 + (y – 40)2 = 1,600
Find the location of the translator and the greatest distance from the main transmitter that the signal can be received.

22. EXAMPLE 4
Radio Translators. The broadcast area of a television station is bounded by the circle
x2 + y2 = 3600, where x and y are measured in miles. A translator station picks up the signal and retransmits it from the center of a circular area bounded by
(x + 30)2 + (y – 40)2 = 1,600
Find the location of the translator and the greatest distance from the main transmitter that the signal can be received.
Strategy Refer to the figure. We will find two distances: the distance from the TV station transmitter to the translator and the distance from the translator to the outer edge of its coverage.
Why The greatest distance of reception from the main transmitter is the sum of those two distances.

23. EXAMPLE 4 Solution Radio Translators.
……Find the location of the translator and the greatest distance from the main transmitter that the signal can be received.
Solution
The coverage of the TV station is bounded by x2 + y2 = 602, a circle centered at the origin with a radius of 60 miles, as shown in yellow in the figure.
Because the translator is at the center of the circle (x + 30)2 + (y – 40)2 = 1600, it is located at (–30, 40), a point 30 miles west and 40 miles north of the TV station. The radius of the translator’s coverage is , or 40 miles.
As shown in the figure, the greatest distance of reception is the sum of d, the distance from the translator to the television station, and 40 miles, the radius of the translator’s coverage.

24. EXAMPLE 4 Solution Radio Translators.
……Find the location of the translator and the greatest distance from the main transmitter that the signal can be received.
Solution
To find d, we use the distance formula to find the distance between the origin, (x1, y1) = (0, 0) and (x2, y2) = (–30, 40).
The distance formula was
introduced in section 9.6.
Substitute for x1, x2, y1 and y2.
Simplify within the radical.
Evaluate the expression within the radical.
Find the square root.
The translator is located 50 miles from the television station, and it broadcasts the signal 40 miles. The greatest reception distance from the main transmitter signal is, therefore, , or 90 miles.

25. Objective 6: Convert the General Form of the Equation of a Parabola to Standard Form to Graph It.
Definition of a Parabola: A parabola is the set of all points in a plane that are equidistant from a fixed point, called the focus, and a fixed line, called the directrix.
Equation of a Parabola
The general forms of the equation of a parabola are:
1. y = ax2 + bx + c The graph opens upward if a > 0 and downward if a < 0
2. x = ay2 + by + c The graph opens to the right if a > 0 and to the left if a < 0

26. Objective 6: Convert the General Form of the Equation of a Parabola to Standard Form to Graph It
The standard form for the equation of a parabola that opens to the right or left is similar to y = a(x – h)2 + k, except that the variables, x and y, exchange positions as do the constants, h and k.
Standard Form of the Equation of a Parabola

27. EXAMPLE 5
Write y = −2x2 + 12x – 13 in standard form and graph it.
Strategy We will complete the square on x to write the equation in standard form, y = a(x – h)2 + k.
Why Standard form contains the expression (x – h)2. We can obtain a perfect-square trinomial that factors into that form by completing the square on x.

28. EXAMPLE 5
Write y = −2x2 + 12x – 13 in standard form and graph it.
Solution
Because the equation is not in standard form, the coordinates of the vertex are not obvious. To write the equation in standard form, we complete the square on x.

29. EXAMPLE 5 Solution continued
Write y = −2x2 + 12x – 13 in standard form and graph it.
Solution continued
This equation is written in the form y = a(x – h)2 + k, where a = –2, h = 3, and k = 5. Thus, the graph of the equation is a parabola that opens downward with vertex at (3, 5) and an axis of symmetry x = 3. We can construct a table of solutions and use symmetry to plot several points on the parabola. Then we draw a smooth curve through the points to get the graph of y = –2x x – 13, as shown below.