Q1 M1 Circles
Q1 M1 Circles
Q1 M1 Circles
SCHOOL
_ _
Before starting this module, I want you to set aside other tasks that will disturb you while enjouing
the lessons. Read the simple instructions below to successfully enjoy the objectives of this kit.
HAVE FUN!
1. Follow carefully all the contents and instructions indicated in every page of this module.
2. Write on your notebook the concepts about the lessons. Writing enhances learning.
3. Practice solving with the help of the given examples. Practice makes perfect.
4. Perform all the provided activities in this module.
5. Let your facilitator/guardian pre-assess your answers.
6. Enjoy studying!
MODULE MAP
CIRCLE PARABOLA
ELLIPSE HYPERBOLA
The graph of the second-degree equation of the form 𝐴𝑥2 + 𝐵𝑥𝑦 + 𝐶𝑦2 + 𝐷𝑥 +
𝐸𝑦 + 𝐹 = 0 is determined by the values of 𝐵2 − 4𝐴𝐶. Table 1.1 provides the type of conic
section given the values of and Conic sections can also be defined in terms of
eccentricity. The table also shows the relationships between eccentricity and the
type of conic sections.
Table 1.1
Graphs of Quadratic Equations
Conic Section Value of 𝐵2 − 4𝐴𝐶 Eccentricity
Circle 2
𝐵 − 4𝐴𝐶 < 0, 𝐵 = 0 𝑜𝑟 𝐴 = 𝐶 𝑒=0
Parabola 𝐵2 − 4𝐴𝐶 = 0 𝑒=1
Ellipse 2
𝐵 − 4𝐴𝐶 < 0, 𝐵 G 0 𝑜𝑟 𝐴 G 𝐶 0<𝑒<1
Hyperbola 𝐵2 − 4𝐴𝐶 > 0 𝑒>1
Example 1: Determine the type of conic section that each general equation will
produce.
Solution:
We will collect all the values of A, B, and C in each equation. Then solve for the
value of 𝐵 2 − 4𝐴𝐶. Interpret the result based on Table 1.1.
OBJECTIVES
INTRODUCTION TO CIRCLES
A circle is a set of all points on a plane that are equidistant from a fixed point on
a plane. The fixed point is called the center, and the distance from the center to any
point of the circle is referred to as the radius. Let say for example we let the center of a
circle be at the fixed point 𝐶(ℎ, 𝑘) and the radius is 𝑟. By applying the distance formula
= √(𝑥2 − 𝑥1)2 + (𝑦2 − 𝑦1)2 , we will arrive at the value of the radius 𝑟 = √(𝑥 − ℎ)2 + (𝑦 − 𝑘)2
or equivalent to 𝑟2 = (𝑥 − ℎ)2 + (𝑦 − 𝑘)2.
The standard form of an equation of the circle is (𝑥 − ℎ)2 + (𝑦 − 𝑘)2 = 𝑟2 with the
radius 𝑟 and center (ℎ, 𝑘), while its standard form radius 𝑟 and center at the origin is
𝑥2 + 𝑦2 = 𝑟2.
This section will discuss how to set up the graph of a circle and determine its
radius. The following examples will illustrate the following sets of problems.
Solution:
Note that 𝑥2 + 𝑦2 = 16 is the standard equation of the circle with center at the
origin (0,0) and radius. We can represent the equation into 𝑥2 + 𝑦2 = 𝑟2, to determine
the radius 𝑟.
𝑥2 + 𝑦2 = 42, thus 𝑟 = 4.
To be able to sketch the graph of the circle, we take all the points that are 4
units from (0,0) to all direction along the plane. The circle is reflected in Figure 1.4.
Figure 1. 4
Example 2: Determine the general equation of the circle whose center is (3, −1) and
whose graph contains the point (7, −1). Sketch the graph.
Solution:
We need to solve for the length of the radius.
We let (ℎ, 𝑘) = (3, −1) and (𝑥, 𝑦) = (7, −1) . (See figure 1.5)
𝑟 = √(𝑥 − ℎ)2 + (𝑦 − 𝑘)2
𝑟 = √(7 − 3)2 + [−1 − (−1)]2
𝑟 = √42 + 02
𝑟 = √16
𝑟=4
(𝑥 − ℎ)2 + (𝑦 − 𝑘)2 = 𝑟2
(𝑥 − 3)2 + [𝑦 − (−1)]2 =
42
(𝑥 − 3)2 + (𝑦 + 1)2 = 16
𝑥2 − 6𝑥 + 9 + 𝑦2 + 2𝑦 + 1 − 16 = 0
𝑥2 + 𝑦2 − 6𝑥 + 2𝑦 − 6 = 0
Figure 1.5
Example 3: Find the general equation of the circle whose center is (2, 6) and whose
radius is 3. Graph the circle.
Solution:
We let (ℎ, 𝑘) = (2, 6) and = 3 . We will make use of the same formula, to establish the
equation of the circle. (See figure 1.6)
(𝑥 − ℎ)2 + (𝑦 − 𝑘)2 = 𝑟2
(𝑥 − 2)2 + (𝑦 − 6)2 = 32
(𝑥 − 2)2 + (𝑦 − 6)2 = 9
𝑥2 − 4𝑥 + 4 + 𝑦2 − 12𝑦 + 36 − 9 = 0
𝑥2 + 𝑦2 − 4𝑥 − 12𝑦 ± 31 = 0
Figure 1.6
The standard form of the equation of the circle where the center (ℎ, 𝑘) and the
radius is 𝑟 is (𝑥 − ℎ)2 + (𝑦 − 𝑘)2 = 𝑟2 . The standard form of the circle is more convenient
in the sense that we can easily identify the center and the radius of a circle.
Another way to represent the equation of a circle, which is called the general
form, is 𝑥2 + 𝑦2 + 𝐷𝑥 + 𝐸𝑦 + 𝐹 = 0. We can convert the equation of a circle in general
form to standard form by completing the squares, the equation may be presented as
𝑥2 + 𝑦2 + 𝐷𝑥 + 𝐸𝑦 + 𝐹 = 0
𝑥2 + 𝐷𝑥 + 𝑌2 + 𝐸𝑦 + 𝐹 = 0
−𝐹
𝐷2 𝐸 2 𝐷 2
𝐸 2
[𝑥 + 𝐷𝑥 + ( ) ] + [𝑦2 + 𝐸𝑦 + ( ) ] = ( )
2
+( ) −𝐹
2 2 2 2
2 2 2 2
(𝑥 + 𝐷) + (𝑦 + 𝐸) = 𝐷 + 𝐸 − 𝐹
2 2 4 4
−
4
�
�
�
�
�
�
�
�
=
√
4
Example 1: Change the equation 𝑥2 + 𝑦2 +
8𝑥 − 6𝑦 = 0 to standard form and determine
the center and the radius of the circle. Sketch
the graph.
Solution:
𝑥2 + 𝑦2 + 8𝑥 − 6𝑦 = 0
(𝑥2 + 8𝑥) + (𝑦2 − 6𝑦) = 0
(𝑥 + 8𝑥 + 16) + (𝑦2 − 6𝑦 + 9) = 16 + 9
2
(𝑥 + 4)2 + (𝑦 − 3)2 = 25
Figure 1.7
Solution:
In order to convert the general equation we will apply completing the square and
factoring the square of binomial.
𝑥2 + 𝑦2 − 6𝑥 + 4𝑦 + 4 = 0
(𝑥2 − 6𝑥) + (𝑦2 + 4𝑦) = −4
(𝑥 − 6𝑥 + 9) + (𝑦2 + 4𝑦 + 4) = −4 + 9 + 4
2
(𝑥 − 3)2 + (𝑦 − 2)2 = 9
𝑟 = 3.
Figure 1.8
Solution:
𝑥2 + 𝑦2 − 4𝑥 − 8𝑦 + 20 = 0
(𝑥2 + 4𝑥) + (𝑦2 − 8𝑦) = −20
(𝑥2 + 4𝑥 + 4) + (𝑦2 − 8𝑦 + 16) = −20 + 4 + 16
(𝑥 + 2)2 + (𝑦 − 4)2 = 0
Figure 1.9
Example 4: Express 𝑥2 + 𝑦2 + 6𝑥 − 10𝑦 + 40 = 0 to standard form.
Solution:
𝑥2 + 𝑦2 + 6𝑥 − 10𝑦 + 40 = 0
(𝑥2 + 6𝑥) + (𝑦2 − 10𝑦) = −40
(𝑥2 + 6𝑥 + 9) + (𝑦2 − 10𝑦 + 25) = −40 + 9 + 25
(𝑥 + 3)2 + (𝑦 − 5)2 = −6
Note that the right side of the equation is negative. This implies that there is no point in
the plane that satisfies the equation 𝑥2 + 𝑦2 + 6𝑥 − 10𝑦 + 40 = 0. Therefore, the circle
does not exist.
This section illustrates how to establish the equation of a circle given different
conditions. Three of the possible cases are presented in the succeeding examples.
Example 1: Determine the equation of the circle which passes through the points
𝑃1(−1, 2), 𝑃2(0, 5), 𝑎𝑛𝑑 𝑃3(2, 1). Sketch the graph.
Solution:
Recall that the general form of the equation of a circle is 𝑥2 + 𝑦2 + 𝐷𝑥 + 𝐸𝑦 + 𝐹 = 0 .
We will substitute the coordinates 𝑃1(−1, 2), 𝑃2(0, 5), 𝑎𝑛𝑑 𝑃3(2, 1) to establish the equation
of the circle.
Let 𝑃1(𝑥1, 𝑦1) = (−1, 2)
25 + 5𝐸 + 𝐹 = 0
25 + 5(−6) + 𝐹 = 0
25 − 30 + 𝐹 = 0
−5 + 𝐹 = 0
𝐹=5
Lastly, we substitute 𝐷 = −2, 𝐸 = −6 𝑎𝑛𝑑 𝐹 = 5 to the general equation of a circle:
𝑥2 + 𝑦2 + 𝐷𝑥 + 𝐸𝑦 + 𝐹 = 0
𝑥2 + 𝑦2 − 2𝑥 − 6𝑦 + 5 = 0
Figure 1. 10
Alternative solution:
𝐶𝑃1 = 𝐶𝑃3
√(𝑥1 − ℎ) + (𝑦1 − 𝑘)2 = √(𝑥3 − ℎ)2 + (𝑦3 − 𝑘)2
2
6ℎ − 2𝑘 = 0 Second Equation
3ℎ − 𝑘 = 0
Combine the first and second equation to establish the coordinates of the center.
First Equation ℎ + 3𝑘 = 10 → ℎ + 3𝑘 = 10
Second Equation 3 (3ℎ − 𝑘 = 0) → (+) 9ℎ − 3𝑘 = 0
10ℎ = 10
ℎ=1
Substitute ℎ = 3 in any of the two equations.
3ℎ − 𝑘 = 0 → 3(1) − 𝑘 = 0
3−𝑘 =0
−𝑘 = −3
𝑘=3
Center (ℎ, 𝑘) = (1,3)
After we have established the center of the circle, we solve for the length of the radius
by selecting any point on the circle, let’s say we select 𝑃1(𝑥1, 𝑦1) = (−1, 2).
𝑟 = 𝐶𝑃1 = √(𝑥1 − ℎ)2 + (𝑦1 − 𝑘)2
𝑟 = √(−1 − 1)2 + (2 − 3)2
𝑟 = √(−2)2 + (−1)2
𝑟 = √4 + 1
𝑟 = √5
(𝑥 − ℎ)2 + (𝑦 − 𝑘)2 = 𝑟2
(𝑥 − 1)2 + (𝑦 − 3)2 =
√52 (𝑥 − 1)2 + (𝑦 − 3)2
=5
𝑥 − 2𝑥 + 1 + 𝑦2 − 6𝑦 + 9 − 5 = 0
2
𝑥2 + 𝑦2 − 2𝑥 − 6𝑦 + 5 = 0
Thus, the equation of the circle is 𝑥2 + 𝑦2 − 2𝑥 − 6𝑦 + 5 = 0.
Example 2: Determine the equation of the circle passing through (4, 0) and (3, 5) with a
line 3𝑥 + 2𝑦 − 7 = 0 passing through the center. Sketch the graph.
Solution:
𝑟1 = 𝑟2
√(𝑥 − 𝑥1) + (𝑦 − 𝑦1)2 = √(𝑥 − 𝑥2)2 + (𝑦 − 𝑦2)2
2
𝑥 − 8𝑥 + 𝑦 − 𝑥 + 6𝑥 − 𝑦2 + 10𝑦 = 9 + 25 − 16
2 2 2
−2𝑥 + 10𝑦 = 18
−2𝑥 + 10𝑦 − 18 = 0
𝑥 − 5𝑦 + 9 = 0
3𝑥 + 2𝑦 − 7 = 0 → 3𝑥 + 2𝑦 − 7 = 0
−3(𝑥 − 5𝑦 + 9 = 0) → (+) −3𝑥 + 15𝑦 − 27 = 0
17𝑦 − 34 = 0
17𝑦 = 34
𝑦=2
Substitute 𝑦 = 2 to 3𝑥 + 2𝑦 − 7 = 0.
3𝑥 + 2𝑦 − 7 = 0
3𝑥 + 2(2) − 7 = 0
3𝑥 + 4 − 7 = 0
3𝑥 − 3 = 0
3𝑥 = 3
𝑥=1
Thus, the center of the circle is (1, 2). Then solve for the radius 𝑟 we get
𝑟 = 𝐶𝑃1 = √(𝑥1 − ℎ)2 + (𝑦1 − 𝑘)2
𝑟 = √(4 − 1)2 + (0 − 2)2
𝑟 = √(3)2 + (−2)2
𝑟 = √9 + 4
𝑟 = √13
Substituting the values of ℎ, 𝑘 𝑎𝑛𝑑 𝑟2 to (𝑥 − ℎ)2 + (𝑦 − 𝑘)2 = 𝑟2.
(𝑥 − ℎ)2 + (𝑦 − 𝑘)2 = 𝑟2
(𝑥 − 1)2 + (𝑦 − 2)2 = √132
(𝑥 − 1)2 + (𝑦 − 2)2 = 13
𝑥 − 2𝑥 + 1 + 𝑦2 − 4𝑦 + 4 − 13 = 0
2
𝑥2 + 𝑦2 − 2𝑥 − 4𝑦 − 8 = 0
Thus, the equation of the circle is 𝑥2 + 𝑦2 − 2𝑥 − 4𝑦 − 8 = 0.
Figure 1.11
Example 3: Find the equation of the circle that circumscribe the triangle determined by
the lines 𝑥 = 0, 𝑦 = 0 and 3𝑥 + 4𝑦 − 24 = 0. Sketch the graph.
Solution:
3𝑥 + 4𝑦 − 24 = 0
3(0) + 4𝑦 − 24 = 0
0 + 4𝑦 − 24 = 0
4𝑦 = 24
𝑦=6 The coordinate is 𝑃2(𝑥2, 𝑦2) = (0, 6).
Combining equation𝑦 = 0 and 3𝑥 + 4𝑦 − 24 = 0 to determine the coordinate of 𝑃3, we
get
3𝑥 + 4𝑦 − 24 = 0
3x + 4(0) − 24 = 0
3x + 0 − 24 = 0
3x = 24
𝑥=8 The coordinate is 𝑃3(𝑥3, 𝑦3) = (8, 0).
Substituting, the vertices of the triangle in the general form 𝑥2 + 𝑦2 + 𝐷𝑥 + 𝐸𝑦 + 𝐹 = 0.
If (𝑥, 𝑦) = (0, 0), then
02 + 02 + 𝐷(0) + 𝐸(0) + 𝐹 = 0
𝐹=0
If (𝑥, 𝑦) = (0, 6), then
02 + 62 + 𝐷(0) + 𝐸(6) + 0 = 0
36 + 6𝐸 = 0
6𝐸 = −36
𝐸 = −6
If (𝑥, 𝑦) = (8, 0), then
82 + 02 + 𝐷(8) + (−6)(0) + 0 = 0
64 + 8𝐷 = 0
8𝐷 = −64
𝐷 = −8 Figure 1.12
Substituting 𝐷 = −8, 𝐸 = −6, 𝑎𝑛𝑑 𝐹 = 0 in the general equation of the circle, we get
𝑥2 + 𝑦2 + (−8)𝑥 + (−6)𝑦 + 0 = 0
𝑥2 + 𝑦2 − 8𝑥 − 6𝑦 = 0
TANGENT TO A CIRCLE
A tangent to a circle is a straight line that intersects the circle at exactly one
point referred as the point of tangency. As shown in the figure below, 𝑃(𝑥, 𝑦) is the point
of tangency while the linear equation 𝐴𝑥 + 𝐵𝑦 + 𝐶 = 0 is the tangent line.
There are four main types of problems concerning tangents to circles which
are part of the discussion in this text. First, a tangent line at a given point on the
circle, second is tangent in prescribed direction, third is the inscribe circle in
polygon, and fourth are tangents to a circle from a point outside the circle.
Note: If a line touches the circle in a single point, then it’s a tangent.
If a line touches the circle in two points, then it’s a secant. If a line does not touch the circle, then there is no solution.
Solution:
𝑥2 + 𝑦2 − 8𝑥 − 14𝑦 + 45 = 0
(2𝑦)2 + 𝑦2 − 8(2𝑦) − 14𝑦 + 45 = 0
4𝑦2 + 𝑦2 − 16𝑦 − 14𝑦 + 45 = 0
5𝑦2 − 30𝑦 + 45 = 0
𝑦2 − 6𝑦 + 9 = 0
(𝑦 − 3)(𝑦 − 3) = 0
(𝑦 − 3)2 = 0
𝑦−3=0
𝑦=3
𝑥 = 2𝑦 → 𝑥 = 2(3) → 𝑥=6
Figure 1.14
Thus, the point of tangency is at (6, 3). Since there is only one solution, this shows that
the line 𝑥 − 2𝑦 = 0 just touches the circle in one place and therefore it is a tangent as
shown in Figure 1. 14.
Example 2: Find the equation of the tangent line to the circle 𝑥2 + 𝑦2 + 6𝑥 − 10𝑦 + 17 = 0
at the point (−2, 1). Sketch the graph.
Solution:
Let 𝑃(𝑥1, 𝑦1) = (−2, 1)
We need to determine the center of the circle, we get
𝑥2 + 𝑦2 + 6𝑥 − 10𝑦 + 17 = 0
(𝑥2 + 6𝑥) + (𝑦2 − 10𝑦) = −17
(𝑥 + 6𝑥 + 9) + (𝑦2 − 10𝑦 + 25) = −17 + 9 + 25
2
(𝑥 + 3)2 + (𝑦 − 5)2 = 17
Thus, (ℎ, 𝑘) = (−3, 5).
The gradient (slope) of CP is
𝑦1 − 𝑘 1−5 −4 −4
= = = = −4
𝑚1 = − ℎ −2 − (−3) −2 + 3 1
𝑥 1
Thus, the gradient (slope) of tangent is 1 and the
4
equation of the tangent is
𝑦− 𝑦−1 𝑦−1
𝑦1
𝑚2 = = =
𝑥 − 1 𝑥 − (−2) 𝑥 + 2
𝑥
We equate 𝑚2 and 1, we get
4
1 𝑦−1
=
4 𝑥+2
𝑥 + 2 = 4𝑦 − 4 Figure 1.15
𝑥 + 2 − 4𝑦 + 4 = 0
𝑥 − 4𝑦 + 6 = 0
Therefore, the equation of the tangent line is 𝑥 − 4𝑦 + 6 = 0 as shown in Figure 1.15.
Example 3: Find the equation of the circle with center (4, 0) and touching the line
2𝑥 − 𝑦 − 18 = 0. Sketch the graph.
Solution:
The radius can be obtained by applying the formula for the distance from a line to a
point.
Let 𝐴 = 2, 𝐵 = −1, 𝐶 = −18, and (𝑥, 𝑦) = (4, 0), substituting we get
𝐴𝑥 + 𝐵𝑦 + 𝐶 2(4) + (−1)(0) + 8−0− −10
𝑟= (−18) 18 =
√𝐴2 + 𝐵2 = = √5
√22 + (−1)2 √4 + 1
Thus the required equation of the circle as shown in Figure 1.16 is
−10 2
(𝑥 − 4)2 + (𝑦 − 0)2 = ( )
√5
100
(𝑥 − 4)2 + 𝑦2 =
5
(𝑥 − 4)2 + 𝑦2 = 20
𝑥 − 8𝑥 + 16 + 𝑦2 − 20 = 0
2
𝑥2 + 𝑦2 − 8𝑥 − 4 = 0
Thus, the equation of the circle is 𝑥 + 𝑦2 − 8𝑥 − 4 = 0.
2
Figure 1.16
Solution:
𝑥2 + 𝑦2 − 6𝑥 + 4𝑦 − 12 = 0
(𝑥2 − 6𝑥)+(𝑦2 + 4𝑦) = 12
(𝑥2 − 6𝑥 + 9)+(𝑦2 + 4𝑦 + 4) = 12 + 9 + 4
(𝑥 − 3)2 + (𝑦 + 2)2 = 25
Take note that the line through the center parallel to the given line takes the
equation of the form 3𝑥 + 4𝑦 − 30 = 0. The equation of a line through the center and
parallel to the given line is
The equation is 3𝑥 + 4𝑦 = 1 or 3𝑥 + 4𝑦 − 1 = 0.
The required tangents are parallel to line 3𝑥 + 4𝑦 − 1 = 0 (note that the line 3𝑥 + 4𝑦 − 1 =
0 cuts through the center of the circle) at a distance 5 (radius) from it. Apply the
𝐴𝑥+𝐵𝑦+𝐶
distance from a line to a point formula 𝑑 = where 𝑑 = 𝑟. Note that 𝐴 = 3, 𝐵 =
√𝐴2+𝐵2
4 𝑎𝑛𝑑 𝐶 = −1.
𝐴𝑥 + 𝐵𝑦 + 𝐶 3𝑥 + 4𝑦 − 1 3𝑥 + 4𝑦 − 1 3𝑥 + 4𝑦 − 1 3𝑥 + 4𝑦 − 1
𝑟= = = = =
2 2
√𝐴 + 𝐵 2
√3 + 4 2
√9 + 16 √25 5
3𝑥 + 4𝑦 − 1
𝑟= = ±5
5
3𝑥 + 4𝑦 − 1
=5 3𝑥 + 4𝑦 − 1
5 = −5
5
3𝑥 + 4𝑦 − 1 = 25 3𝑥 + 4𝑦 − 1 = −25
3𝑥 + 4𝑦 − 1 − 25 = 0 3𝑥 + 4𝑦 − 1 + 25 = 0
3𝑥 + 4𝑦 − 26 = 0 3𝑥 + 4𝑦 + 24 = 0
Figure 1.17
Solution:
𝑥 + 2𝑦 − 4 = 0
2𝑦 = −𝑥 + 4
1
𝑦=− 𝑥+2
2
𝑚1 =
1
− and slope of the perpendicular line to 𝑥 + 2𝑦 − 4 = 0 is 𝑚2 = 2. Then we
2
determine the center of the circle 𝑥2 + 𝑦2 − 8𝑥 − 4 = 0.
𝑥2 + 𝑦2 − 8𝑥 − 4 = 0
(𝑥2 − 8𝑥) + 𝑦2 = 4
(𝑥 − 8𝑥 + 16) + 𝑦2 = 4 + 16
2
(𝑥 − 4)2 + 𝑦2 = 20
Thus, (ℎ, 𝑘) = (4, 0) and 𝑟 = √20. Then we apply the formula of a distance from a point to
a line letting (𝑥1, 𝑦1) = (4, 0) and 𝑚2 = 𝑚 = 2, we get
|𝑚𝑥1 + 𝑏 − 𝑦1|
𝑟=
√𝑚2 + 1
|2(4) + 𝑏 − 0|
√20 =
√22 + 1
|8 + 𝑏|
√20 =
√5
(√20)2 = (8 + 𝑏)2
2
√5
20 = (8 + 𝑏)2
5
(8 + 𝑏)2 = 100
√(8 + 𝑏)2 = √100 Figure 1.18
8 + 𝑏 = ±10
8 + 𝑏 = 10 8 + 𝑏 = −10
𝑏 = 10 − 8 𝑏 = −10 − 8
𝑏=2 𝑏 = −18
Substituting the values of 𝑚(slope) and 𝑏(𝑦-intercept) in 𝑦 = 𝑚𝑥 + 𝑏, the equation of the
tangents are 𝑦 = 2𝑥 + 2 (𝑜𝑟 2𝑥 − 𝑦 + 2 = 0) and 𝑦 = 2𝑥 − 18 (𝑜𝑟 2𝑥 − 𝑦 − 18 = 0). See
Figure 1.18
Inscribed circle (or incenter of a triangle) is the largest possible circle that can
be drawn inside the triangle in which each of the triangles’ sides is a tangent to the
circle. The incenter is the point at which the angle bisectors of a triangle intersect and it
is the center of the circle that can be inscribed in a triangle.
Figure 1.19 shows the graph of the circle inscribed in a triangle determined
by the line 𝐿1, 𝐿2 and 𝐿3. 𝐶 is the incenter of the circle.
Figure 1.19
Solution:
Note that the center of the circle is located at the point of intersection of the
bisectors of two angles of the triangle. We can determine the equation of the angle
bisector by
𝐴𝑥+𝐵𝑦+𝐶
applying the formula 𝑑 = , the point being the center of the circle.
√𝐴2+𝐵2
𝐴1𝑥 + 𝐵1𝑦 + 2𝑥 − 𝑦 + 0 2𝑥 − 2𝑥 − 𝑦
𝐶1 𝑦
𝑟1 = = = =
√𝐴 2 + 𝐵 2
√22 + (−1)2 √4 + 1 √5
1 1
𝐴2𝑥 + 𝐵2𝑦 + 𝐶2 2𝑥 + 𝑦 − 16 2𝑥 + 𝑦 − 16 2𝑥 − 𝑦 − 16
𝑟2 = = = =
√𝐴22 + 𝐵22 √22 + 12 √4 + 1 √5
𝐴3𝑥 + 𝐵3𝑦 + 𝐶3 𝑥 − 2𝑦 − 9 𝑥 − 2𝑦 − 9 𝑥 − 2𝑦 − 9
𝑟3 =
= = =
√𝐴32 2
√1 + (−2) 2
√1 + 4 √5
+ 𝐵32
Equating −𝑟1 = 𝑟2 (angle bisector of 𝐿1 and 𝐿2) and 𝑟2 = 𝑟3 (angle bisector of 𝐿2 and 𝐿3,
we get
−𝑟1 = 𝑟2
2𝑥 − 2𝑥 + 𝑦 − 16
𝑦 =
− √5
√5
−2𝑥 + 𝑦 = 2𝑥 + 𝑦 − 16
−2𝑥 − 𝑦 − 2𝑥 − 𝑦 + 16 = 0
−4𝑥 + 16 = 0
−4𝑥 = −16
𝑥=4
𝑟2 = 𝑟3
2𝑥 + 𝑦 − 16
= 𝑥 − 2𝑦 − 9
√5 √5
2𝑥 + 𝑦 − 16 = 𝑥 − 2𝑦 − 9
2𝑥 + 𝑦 − 16 − 𝑥 + 2𝑦 + 9 = 0
𝑥 + 3y − 7 = 0
The equation of the angle bisector of 𝐿1 and 𝐿2 is 𝑥 = 4, while the equation of the angle
bisector of 𝐿2 and 𝐿3 is 𝑥 + 3y − 7 = 0. Solving for the center (ℎ, 𝑘), we get
𝑥 + 3𝑦 − 7 = 0
4 + 3𝑦 − 7 = 0
3𝑦 − 3 = 0
3𝑦 = 3
𝑦=1
Thus, the center of the circle is 𝐶(ℎ, 𝑘) = (4, 1). Substituting the (4, 1) to any of the 𝑟1,𝑟2
and 𝑟3 yields the radius of the circle, we get
2𝑥 − 2(4) − 1 8 − 1 7
𝑟1 = = = =
𝑦 √5 √5 √5
√5
Therefore, the equation of the circle tangent to the three given lines is
5𝑥2 + 5𝑦2 − 40𝑥 − 10𝑦 + 36 = 0, as shown in Figure 1.20.
Figure 1.20
Solution:
Let 𝑃1(𝑥1, 𝑦1) = (1, 7) and 𝐶1: 𝑥2 + 𝑦2 + 4𝑥 − 6𝑦 + 4 = 0 .
We solve for the center (ℎ, 𝑘) of the circle 𝐶1.
𝑥2 + 𝑦2 + 4𝑥 − 6𝑦 + 4 = 0
(𝑥2 + 4𝑥) + (𝑦2 − 6𝑦) = −4
(𝑥 + 4𝑥 + 4) + (𝑦2 − 6𝑦 + 9) = −4 + 4 + 9
2
(𝑥 + 2)2 + (𝑦 − 3)2 = 9
𝐶𝑃1 = √(𝑥1 − ℎ)2 + (𝑦1 − 𝑘)2 = √(1 + 2)2 + (7 − 3)2 = √32 + 42 = √9 + 16 = √25 = 5
𝐶𝑃2 = 3
𝐶𝑃3 = √(𝐶𝑃1)2 − (𝐶𝑃2)2 = √52 − 32 = √25 − 9 = √16 = 4
16 (𝑥 − 1)2 + (𝑦 − 7)2 = 16
𝑥 − 2𝑥 + 1 + 𝑦2 − 14𝑦 + 49 = 16
2
𝑥 + 𝑦2 − 2𝑥 − 14𝑦 + 1 + 49 − 16 = 0
2
𝑥2 + 𝑦2 − 2𝑥 − 14𝑦 + 34 = 0
𝐶1: 𝑥2 + 𝑦2 + 4𝑥 − 6𝑦 + 4 = 0
𝐶2: (𝑥 − 1)2 + (𝑦 − 7)2 = 16
𝐶1 − 𝐶2 = 0
𝑥2 + 𝑦2 + 4𝑥 − 6𝑦 + 4 − (𝑥2 + 𝑦2 − 2𝑥 − 14𝑦 + 34) = 0
𝑥2 + 𝑦2 + 4𝑥 − 6𝑦 + 4 − 𝑥2 − 𝑦2 + 2𝑥 + 14𝑦 − 34 = 0
6𝑥 + 8𝑦 − 30 = 0
3𝑥 + 4𝑦 − 15 = 0
3𝑥 = −4𝑦 + 15
−4𝑦 + 15
𝑥=
3
−4𝑦 + 15 𝑥2 + 𝑦2 + 4𝑥 − 6𝑦 + 4 = 0
2 −4𝑦 + 15
( ) + 𝑦2 + 4 ( ) − 6𝑦 + 4 = 0
3 3
16𝑦2 − 120𝑦 + 225 −16𝑦 + 60
+ 𝑦2 + − 6𝑦 + 4 = 0
9 3
16𝑦2 − 120𝑦 + 225 + 9𝑦2 − 48𝑦 + 180 − 54𝑦 + 36 = 0
25𝑦2 + 9𝑦2 − 222𝑦 + 441 = 0
(25𝑦 − 147)(𝑦 − 3) = 0
Solve for 𝑦,
25𝑦 − 147 = 0 𝑦−3=0
25𝑦 = 147 𝑦=3
147
𝑦=
25
147 −4𝑦+15
Substitute 𝑦 = to 𝑥 = , we get
25 3
147
147 −4( 25 )+15 71 71
If 𝑦= , then =− and
147
the coordinate is (− , ) or −2.84, 5.88).
25 3 25 25 25
−4(3)+15
If 𝑦 = 3, then =1 and the coordinate is (1, 3).
3
𝑦1 − 𝑦 147
𝑚 = 7 − 25 = 7
=
71
1
𝑥1 − 𝑥 1 − (− ) 24
25
𝑦1 − 𝑦 7 − 3 4
= =
𝑚2 = − 𝑥 1 − 1 0
𝑥 1
7 4
𝑦−7 = (𝑥 − 1) 𝑦−7= (𝑥 − 1)
24 0
24(𝑦 − 7) = 7(𝑥 − 1) 0(𝑦 − 7) = 4(𝑥 − 1)
24𝑦 − 168 = 7𝑥 − 7 0 = 4𝑥 − 4
−7𝑥 + 24𝑦 − 168 + 7 = 0 − 4𝑥 = −4
−7𝑥 + 24𝑦 − 161 = 0 𝑥=1
7𝑥 − 24𝑦 + 161 = 0 𝑥−1=0
UNIT EXERCISES
A. Determine the type of conic section that each general equation will produce.
1. 𝑥2 + 3𝑦2 + 4𝑥 − 6𝑦 + 1 = 0 4. 2𝑥2 + 2𝑦2 − 12𝑥 + 2𝑦 + 1 = 0
2. 𝑥2 + 𝑦2 − 2𝑥 − 4𝑦 + 1 = 0 5. 𝑥2 − 8𝑥 − 8𝑦 − 24 = 0
CONGRATULATIONS!
References:
Pre-Calculus, Debbie Marie B. Verzosa, Ph.D., Richard B. Eden, Ph.D., and Ian June L.
Garces, Ph.D., Vibal Group, Inc., 2016