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PRE-CAL-1 - hehe
Engineering (Eng.1221)
AMA Computer Learning Center
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ACLC COLLEGE OF TACLOBAN
Tacloban City
SENIOR HIGH SCHOOL DEPARTMENT
Learning Outcomes of the Lesson
At the end of the lesson, the student is able to:
(1) illustrate the different types of conic sections: parabola, ellipse, circle, hyperbola, and degenerate cases;
(2) define a circle;
(3) determine the standard form of equation of a circle;
(4) graph a circle in a rectangular coordinate system; and
(5) solve situational problems involving conic sections (circles).
Lesson Outline
(1) Introduction of the four conic sections, along with the degenerate conics
(2) Definition of a circle
(3) Derivation of the standard equation of a circle
(4) Graphing circles
(5) Solving situational problems involving circles
Introduction
We introduce the conic sections, a particular class of curves which sometimes appear in nature and which have
applications in other fields. In this lesson, we discuss the first of their kind, circles. The other conic sections will be
covered in the next lessons.
1.1. An Overview of Conic Sections
We introduce the conic sections (or conics), a particular class of curves which oftentimes appear in nature
and which have applications in other fields. One of the first shapes we learned, a circle, is a conic. When
you throw a ball, the trajectory it takes is a parabola. The orbit taken by each planet around the sun is an
ellipse. Properties of hyperbolas have been used in the design of certain telescopes and navigation
systems. We will discuss circles in this lesson, leaving parabolas, ellipses, and hyperbolas for subsequent
lessons.
Circle (Figure 1) - when the plane is horizontal
Ellipse (Figure 1) - when the (tilted) plane intersects only one cone to form a bounded curve.
Parabola (Figure 1) - when the plane intersects only one cone to form an unbounded curve
Hyperbola (Figure 1) - when the plane (not necessarily vertical) intersects both cones to form two
unbounded curves (each called a branch of the hyperbola).
Lesson 1 Introduction to Conic Sections and Circles
We can draw these conic sections (also called conics) on a rectangular coordinate plane and find their equations. To
be able to do this, we will present equivalent definitions of these conic sections in subsequent sections, and use
these to find the equations. There are other ways for a plane and the cones to intersect, to form what are referred to
as degenerate conics: a point, one line, and two lines. See Figures 1, 1 and 1.
1.1. Definition and Equation of a Circle
A circle may also be considered a special kind of ellipse (for the special case when the tilted plane is
horizontal). For our purposes, we will distinguish between these two conics.
See Figure 1, with the point C(3, 1) shown. From the figure, the distance of A(2, 1) from C is AC = 5. By
the distance formula, the distance of B(6, 5) from C is BC = √( 6 − 3 )
2
+ ( 5 − 1 )
2
= 5 are other
points P such that P C = 5. The collection of all such points which are 5 units away from C, forms a circle.
The term radius is both used to refer to a segment from the center C to a point P on the circle, and the length of this
segment. See Figure 1, where a circle is drawn. It has center C(h, k) and radius r > 0. A point P(x, y) is on the circle
if and only if P C = r. For any such point then, its coordinates should satisfy the following.
P C = r
√(𝑥 − ℎ)
2
+ ( 5 − 1 )
2
= r
(x - h)
2
- (y - k)
2
= r
2
Let C be a given point. The set of all points P having
the same distance from C is called a circle. The point
C is called the center of the circle, and the common
distance its radius.
an equation of the circle in general form.
If the equation of a circle is given in the general form
Ax
2
- Ay
2
- Cx + Dy + E = 0, A ≠ 0,
x
2
- y
2
- Cx + Dy + E = 0,
we can determine the standard form by completing the square in both variables.
Completing the square in an expression like x
2
- 14x means determining the term to be added that will
produce a perfect polynomial square. Since the coefficient of x
2
is already 1, we take half the coefficient of x
and square it, and we get 49. Indeed, x
2
- 14x + 49 = (x + 7)
2
is a perfect square. To complete the square
in, say, 3x
2
- 18x, we factor the coefficient of x
2
from the expression: 3(x
2
- 6x), then add 9 inside. When
completing a square in an equation, any extra term introduced on one side should also be added to the
other side
Example 1.1. Identify the center and radius of the circle with the given equation in each item. Sketch its
graph, and indicate the center.
(1) x
2
- y
2
- 6x = 7
(2) x
2
- y
2
- 14x + 2y = 14
(3) 16x
2
- 16y
2
- 96x - 40y = 315
Solution. The first step is to rewrite each equation in standard form by completing the square in x and in y.
From the standard equation, we can determine the center and radius.
(1) x
2
- 6x + y
2
= 7
x
2
- 6x +9+ y
2
=7+
(x - 3)
2
- y
2
= 16
Center (3, 0), r = 4, Figure 1.
(2) x
2
- 14x + y
2
- 2y = - 14
x
2
- 14x + 49 + y
2
- 2y +1 = - 14 + 49 + 1
(x - 7)
2
- (y + 1)
2
= 36
Center (7, - 1), r = 6, Figure 1.
(3) 16x
2
- 96x + 16y
2
- 40y = 315
16(x
2
- 6x) + 16 (y
2
-
5
2
𝑦) = 315
16(x
2
- 6x + 9) + 16(y
2
-
5
2
𝑦 +
25
16
) = 315 + 16(9) + 16(
25
16
)
16(x + 3)
2
- 16(y
2
-
5
4
)
2
= 484
(x + 3)
2
- (y -
5
4
)
2
=
484
16
=
121
4
= (
11
2
)
2
Center (-3,
5
4
) , r = 5, Figure 1.
In the standard equation (x - h)
2
- (y - k)
2
= r
2
, both the two squared terms on the left side have coefficient 1. This is
the reason why in the preceding example, we divided by 16 at the last equation.
1.1. Situational Problems Involving Circles
We now consider some situational problems involving circles.
Example 1.1. A street with two lanes, each 10 ft wide, goes through a semicircular tunnel with radius 12 ft. How
high is the tunnel at the edge of each lane? Round o↵ to 2 decimal places.
Figure 1.
Solution. We draw a coordinate system with origin at the middle of the highway, as shown in Figure 1. Because
of the given radius, the tunnel’s boundary is on the circle x
2
- y
2
= 12
2
. Point P is the point on the arc just above the
edge of a lane, so its x-coordinate is 10. We need its y-coordinate. We then solve 10
2
- y
2
= 12
2
for y > 0, giving us y
=
√
11 ≈ 6. 63 ft.
Example 1.1. A piece of a broken plate was dug up in an archaeological site. It was put on top of a grid, as shown
in Figure 1, with the arc of the plate passing through A(7, 0), B(1, 4) and C(7, 2). Find its center, and the standard
equation of the circle describing the boundary of the plate.
Solution. We first determine the center. It is the intersection of the perpendicular bisectors of AB and BC (see Figure
1). Recall that, in a circle, the perpendicular bisector of any chord passes through the center. Since the midpoint
M of AB is (
− 7 + 1
2
,
0 + 4
2
) = (-3, 2) and m AB
=
4 + 0
1 + 7
=
1
2
, the perpendicular bisector of AB has equation y - 2 = 2(x + 3), or
equivalently, y = 2x - 4.
Since the midpoint N of BC is (
1 + 7
2
,
4 + 2
2
)= (4, 3), and mBC =
2 − 4
7 − 1
=
− 1
3
, the perpendicular bisector of BC has
equation y - 3 = 3(x - 4), or equivalently, y = 3x - 9.
The intersection of the two lines y = 2x - 4 and y = 3x - 9 is (1, - 6) (by solving a system of linear equations). We can
take the radius as the distance of this point from any of A, B or C (it’s most convenient to use B in this case). We then
get r = 10. The standard equation is thus (x - 1)
2
- (y + 6)
2
= 100.
PRE-CAL-1 - hehe
Course: Engineering (Eng.1221)
University: AMA Computer Learning Center
