Section 22–2
◆
625
The Circle
Dollars
y
Purchase
price
P
Book value y = f(t)
Salvage
value
S
Years
t
Useful life L
FIGURE 22–32
Straight-line depreciation.
22–2 The Circle
Definition
Anyone who has drawn a circle using a compass will not be surprised by the following definition of the circle:
Definition
of a Circle
A circle is a plane curve all points of which are at
a fixed distance (the radius) from a fixed point (the
centre).
303
Circle with Centre at the Origin
We will now derive an equation for a circle of radius r, starting with the simplest case of a
circle whose centre is at the origin (Fig. 22–33). Let x and y be the coordinates of any point P
on the circle. The equation we develop will give a relationship between x and y that will have
two meanings: geometrically, (x, y) will represent a point on the circle; and algebraically, the
numbers corresponding to those points will satisfy that equation.
We observe that, by the definition of a circle, the distance OP must be constant and equal
to r. But, by the distance formula (Eq. 283),
y
r
P(x, y)
r
O
r
x
OP x2 y2 r
Squaring, we get a standard equation of a circle (also called standard form of the equation).
Standard Equation,
Circle of Radius r:
Centre at Origin (O)
x2 y2 r2
Note that both x2 and y2 have the
same coefficient. Otherwise, the
graph is not a circle.
304
Example 21: Write, in standard form, the equation of a circle of radius 3, whose centre is
at the origin.
◆◆◆
Solution:
x2 y2 32 9
FIGURE 22–33 Circle with
centre at origin.
◆◆◆
626
y
Chapter 22
C(h, k)
r
P(x, y)
k
0
h
◆
Analytic Geometry
Circle with Centre Not at the Origin
Figure 22–34 shows a circle whose centre has the coordinates (h, k). We can think of the difference between this circle and the one in Fig. 22–33 as having its centre moved or translated h
units to the right and k units upward. Our derivation is similar to the preceding one.
CP r (x h)2 (y k)2
x
FIGURE 22-34 Circle with
centre at (h, k).
Squaring, we get the following equation:
Standard Equation,
Circle of Radius r:
Centre at (h, k)
(x h)2 (y k)2 r2
305
◆◆◆ Example 22: Write, in standard form, the equation of a circle of radius 5 whose centre is
at (3, 2).
Solution: We substitute into Eq. 305 with r 5, h 3, and k 2.
(x 3)2 [y (2)]2 52
(x 3)2 (y 2)2 25
◆◆◆
Example 23: Find the radius and the coordinates of the centre of the circle (x 5)2 (y 3)2 16.
◆◆◆
Solution: We see that r2 16, so the radius r is 4. Also, since
xhx5
then
h 5
and since
yky3
then
k3
So the centre is at (5, 3) as shown in Fig. 22–35.
y
6
(x + 5)2 + (y − 3)2 = 16
4
C(−5, 3)
2
−6
−4
−2
FIGURE 22–35
0
x
◆◆◆
Section 22–2
◆
627
The Circle
Common
Error
It is easy to get the signs of h and k wrong. In Example 23,
do not take
h 5
and
k 5
Translation of Axes
We see that Eq. 305 for a circle with its centre at (h, k) is almost identical to Eq. 304 for a circle
with centre at the origin, except that x has been replaced by (x h) and y has been replaced
by (y k). This same substitution will, of course, work for curves other than the circle, and we
will use it to translate or shift the axes for the other conic sections.
Translation
of Axes
To translate or shift the axes of a curve to the left by a
distance h and downward by a distance k, replace x by
(x h) and y by (y k) in the equation of the curve.
298
General Equation of a Circle
A second-degree equation in x and y which has all possible terms would have an x2 term, a y2
term, an xy term, and all terms of lesser degree as well. The general second-degree equation
is usually written with terms in the following order, where A, B, C, D, E, and F are constants:
General SecondDegree Equation
Ax Bxy Cy Dx Ey F 0
2
2
297
If we expand Eq. 305, we get
(x h)2 (y k)2 r2
x2 2hx h2 y2 2ky k2 r2
Rearranging gives
x2 y2 2hx 2ky (h2 k2 r2) 0
or an equation with D, E, and F as constants.
General Equation
of a Circle
x2 y2 Dx Ey F 0
306
Comparing this with the general second-degree equation, we see that the general second-degree
equation
Ax2 Bxy Cy2 Dx Ey F 0
represents a circle if B 0 and A C.
◆◆◆
Example 24: Write the equation of Example 23 in general form.
Solution: The equation, in standard form, was
(x 5)2 (y 3)2 16
There are six constants in
this equation, but only five are
independent. We can divide
through by any constant and thus
make it equal to 1.
628
Chapter 22
◆
Analytic Geometry
Expanding, we obtain
x2 10x 25 y2 6y 9 16
or
x2 y2 10x 6y 18 0
◆◆◆
Changing from General to Standard Form
When we want to go from general form to standard form, we must complete the square, both
for x and for y.
Example 25: Write the equation 2x2 2y2 18x 16y 60 0 in standard form. Find
the radius and centre, and plot the curve.
◆◆◆
We learned how to complete the
square in Sec. 14–2 of Chapter
14, “Quadratic Equations.”
Glance back there if you need to
refresh your memory.
Solution: We first divide by 2:
x2 y2 9x 8y 30 0
and separate the x and y terms:
(x2 9x) (y2 8y) 30
81
We complete the square for the x terms with because we take 12 of (9) and square it.
4
2
Similarly, for completing the square for the y terms, 12 8 16. Completing the square on
the left and compensating on the right gives
p
y
0
2
4
6
8
p 2
81
81
x 9x q (y2 8y 16) 30 16
4
4
x
−2
−4
Factoring each group of terms yields
−6
−8
q
(
C
9
,
2
p
)
−4
FIGURE 22–36.
9 2
5 2
x q (y 4)2 p q
2
2
So
5
r
2
9
h
2
k 4
The circle is shown in Fig. 22–36.
◆◆◆
Applications
◆◆◆ Example 26: Write an equation for the circle shown in Fig. 22–37, and find the dimensions A and B produced by the circular cutting tool.
Estimate: The easiest way to get an estimate here is to make a sketch and measure the distances. We get
A 2.6 cm
Note that the orientations of the
x and y axes are reversed in this
example.
and
B 2.7 cm
Solution: In order to write an equation, we must have coordinate axes. Otherwise, the quantities x and y in the equation will have no meaning. Since no axes are given, we are free to draw
them anywhere we please. Our first impulse might be to place the origin at the corner p, since
most of the dimensions are referenced from that point, but our equation will be simpler if we
Section 22–2
◆
629
The Circle
place the origin at the centre of the circle. Let us draw axes as shown, so that not all our numbers
will be negative. The equation of the circle is then
x2 y2 (1.500)2
2.2500
r
0
y
0
.50
=1
cm
3.715 cm
x
y
A
4.155 cm
3.148 cm
B
p
4.146 cm
x
FIGURE 22–37
The distance from the y axis to the top edge of the block is 4.155 3.148 1.007 cm.
Substituting this value for x into the equation of the circle will give the corresponding value for
y. When x 1.007,
y2 2.250 (1.007)2 1.236
y 1.112 cm
Similarly, the distance from the x axis to the right edge of the block is 4.146 3.715 0.431
cm. When y 0.431,
x2 2.250 (0.431)2 2.064
x 1.437
Finally,
A 3.715 1.112 2.603 cm
and
B 4.155 1.437 2.718 cm
Exercise
◆
The Circle
Equation of a Circle
Write the equation of each circle in standard form.
1. centre at (0, 0); radius 7
2. centre at (0, 0); radius 4.82
◆◆◆
630
Chapter 22
3.
4.
5.
6.
◆
Analytic Geometry
centre at (2, 3); radius 5
centre at (5, 2); radius 10
centre at (5, 3); radius 4
centre at (3, 2); radius 11
Find the centre and radius of each circle.
7. x2 y2 49
8. x2 y2 64.8
9. (x 2)2 (y 4)2 16
10. (x 5)2 (y 2)2 49
11. (y 5)2 (x 3)2 36
12. (x 2.22)2 (y 7.16)2 5.93
13. x2 y2 8x 0
14. x2 y2 2x 4y 0
15. x2 y2 10x 12y 25 0
16. x2 y2 4x 2y 36
17. x2 y2 6x 2y 15
18. x2 y2 2x 6y 39
Write the equation of each circle in general form.
19. centre at (1, 1); passes through (4, 3)
20. diameter joins the points (2, 5) and (6, 1)
Tangent to a Circle
Write the equation of the tangent to each circle at the given point. (Hint: The slope of the tangent is the negative reciprocal of the slope of the radius to the given point.)
21. x2 y2 25 at (4, 3)
22. (x 5)2 (y 6)2 100 at (11, 14)
23. x2 y2 10y 0 at (4, 2)
Intercepts
Find the x and y intercepts for each circle. (Hint: Set x and y, in turn, equal to zero to find the
intercepts.)
24. x2 y2 6x 4y 4 0
25. x2 y2 5x 7y 6 0
Intersecting Circles
Find the point(s) of intersection. (Hint: Solve each pair of equations simultaneously as we did
in Exercise 8 in Chapter 14.)
26. x2 y2 10x 0 and x2 y2 2x 6y 0
27. x2 y2 3y 4 0 and x2 y2 2x 5y 2 0
28. x2 y2 2x 6y 2 0 and x2 y2 4x 2 0
Applications
Even though you might be able
to solve some of these with only
the Pythagorean theorem, we
suggest that you use analytic
geometry for the practice.
29. Prove that any angle inscribed in a semicircle is a right angle.
30. Write the equation of the circle in Fig. 22–38, taking the axes as shown. Use your equation
to find A and B.
Section 22–2
◆
631
The Circle
y
y
B
P
1.8
0
m
cm
10
m
1.
2.
50
6.25 cm
h
1.80 m
1.98 cm
1.30 m
0 1.22
cm
A
x
x
0
8.75 cm
2.20 m
FIGURE 22–39
FIGURE 22–38
31. Write the equations for each of the circular arches in Fig. 22–39, taking the axes as shown.
Solve simultaneously to get the point of intersection P, and compute the height h of the
column.
32. Write the equation of the centreline of the circular street shown in Fig. 22–40, taking the
origin at the intersection O. Use your equation to find the distance y.
33. Each side of a Gothic arch is a portion of a circle. Write the equation of one side of the
Gothic arch shown in Fig. 22–41. Use the equation to find the width w of the arch at a
height of 3.00 m.
y
y
0m
20
y
35.0 m
w
x
0
10
.0
3.00 m
m
x
0
85.0 m
FIGURE 22–40
90.0 m
Circular street.
6.00 m
FIGURE 22–41 Gothic arch.
632
Chapter 22
◆
Analytic Geometry
22–3 The Parabola
Directrix
L
9.0 cm
Definition
A parabola has the following definition:
P2
Definition
of a
Parabola
9.0 cm
V
F
A parabola is the set of points in a plane, each of
which is equidistant from a fixed point, the focus,
and a fixed line, the directrix.
307
6.0 cm
Example 27: Use Definition 307 above to construct a parabola for which the distance from
focus to directrix is 6.0 cm.
◆◆◆
P1
Solution: We draw a line to represent the directrix as shown in Fig. 22–42, and we indicate a
focus 6.0 cm from that line. Then we draw a line L parallel to the directrix at some arbitrary
distance, say, 9.0 cm. With that same 9.0-cm distance as radius and F as centre, we use a comFIGURE 22–42 Construction pass to draw arcs intersecting L at P1 and P2. Each of these points is now at the same distance
of a parabola.
(9.0 cm) from F and from the directrix, and is hence a point on the parabola. Repeat the con◆◆◆
struction with distances other than 9.0 cm to get more points on the parabola.
Figure 22–43 shows the typical shape of a parabola. The parabola has an axis of symmetry
which intersects it at the vertex. The distance p from directrix to vertex is equal to the directed
distance from the vertex to the focus.
Directrix
Vertex
p
Axis of symmetry
p
Focus
FIGURE 22–43 Parabola.
◆◆◆
Standard Equation of a Parabola with Vertex at the Origin
Let us place the parabola on coordinate axes with the vertex at the origin and with the axis of
symmetry along the x axis, as shown in Fig. 22–44. Choose any point P on the parabola. Then,
by the definition of a parabola, FP AP. But in right triangle FBP,
FP (x p)2 y2
and
AP p x
But, since FP AP,
(x p)2 y2 p x
Squaring both sides yields
(x p)2 y2 p2 2px x2
x2 2px p2 y2 p2 2px x2