1
Section 1.4 – Circles
Objective #1: Writing the Equation of a Circle in Standard Form.
We begin by giving a definition of a circle:
Definition: A Circle is the set of all points that are equidistant from a fixed
point (h, k) called the center in the xy-plane. The radius of the circle, r, is
the length between the center to a point on the circle.
(x, y)
To derive the equation of a circle, let (x, y)
represent any point on the circle. To find the
r
distance between this point and the center in
the xy-plane, we can use the distance formula:
(h, k)
d=
(x 2 − x1)2 +(y 2 − y1)2
But, since (x1, y1) = (h, k), (x2, y2) = (x, y) and d = r, we get:
r€=
(x −h)2 +(y −k)2
Squaring both sides and rearranging yields the standard form of the
equation of a circle:
(x – h)2 + (y – k)2 = r2
€
If we were to expand, combine like terms, and get zero on one side of the
equation, this would produce the general form of the equation of a circle.
Definition: Given a circle of radius r centered at (h, k), then
1) The standard form of the equation of the circle is:
(x – h)2 + (y – k)2 = r2
Note that if the origin (0, 0) is the center, then the equation becomes
x2 + y 2 = r2
2) The general form of the equation of the circle is:
x2 + y2 + ax + by + c = 0 where a, b, and c are constants.
3) The unit circle is a circle with radius 1 centered at the origin:
x2 + y 2 = 1
Given the radius and the center, write a) the standard form and b)
general form of each equation:
Ex. 1 r = 7, (h, k) = (– 3, 2)
2
Solution:
a) Replace r with 7 and (h, k) with (– 3, 2) in the standard form:
(x – h)2 + (y – k)2 = r2
(x – (– 3))2 + (y – (2))2 = 72 (simplify)
(x + 3)2 + (y – 2)2 = 49 Standard form
b) To find the general form, expand, combine like terms, and get zero
on one side: (x + 3)2 + (y – 2)2 = 49
(x + 3)(x + 3) + (y – 2)(y – 2) = 49 (FOIL)
x2 + 6x + 9 + y2 – 4y + 4 = 49
(combine like terms)
2
2
x + y + 6x – 4y + 13 = 49
(subtract 49 from both sides)
2
2
x + y + 6x – 4y – 36 = 0 General form
Ex. 2 r =
5
4
1
2
, (h, k) = ( , –
Solution:
a) Replace r with
5
4
7
8
)
1
2
and (h, k) with ( , –
(x €
– h)2 €
+ (y – k)2 = r2
€
(x
(x
1
7
– ( ))2 + (y – (– ))2 =
2
8
€1 2
€25
7
– ) + (y + )2 =
2
8
16
2
( )
5
4
7
8
) in the standard form:
(simplify)
€
Standard form
€
€ expand, combine like terms, and get zero
b)€To find the general
form,
7 2
25
) =
8
16
1
1
7
7
25
(x – )(x – ) + (y + )(y + ) =
(FOIL)
2
2
8
8
16
1
7
49
25
x2 – x + + y 2 + y +
=
(combine like terms)
4
64
16
€4
€
€
7
65
25
25
x2 + y 2 – x + y +
=
(subtract
from both
4
64
16
16
€
€
€
€
7
35
x2 + y 2 – x + y –
= 0 General form
4€
64
€
€
€
€on one side:
€
€
(x
€–
1 2
)
2
€
€ a Circle
€
Objective #2: Graphing
+ (y +
sides)
€
€ the €
If the equation of
circle is in standard form, we simply determine the
center (h, k) and the radius r. We then locate the center of the circle on the
graph paper and plot four points r units above, below, to the left, and to the
right of the center. Finally, we draw a circle through those four points.
3
In the case where the equation is not in standard form, we will first need to
rewrite the equation in standard form. This may include the need to
complete the square in order to get the equation in standard form.
Find a) the center and the radius of the circle, b) sketch the graph of
the circle, and c) find the intercepts:
Ex. 3
6(x – 2)2 + 6(y + 1)2 = 54
Solution:
a) To get the equation in standard
form, divide both sides by 6. This
will yield: (x – 2)2 + (y + 1)2 = 9
Thus, (h, k) = (2, – 1) and r = 3.
b) To graph the circle, we locate
the center (2, – 1) and then plot
four points that are 3 units above,
below, to the left, and to the right
of (2, – 1). Finally, we draw a circle
through those four points (see figure
on the right).
c) To find the x-intercepts, let y = 0 and solve:
(x – 2)2 + (0 + 1)2 = 9
(x – 2)2 + 1 = 9
(subtract 1 from both sides)
2
(x – 2) = 8
(use the square root property)
x–2=± 8
(add 2 to both sides and simplify)
x = 2 ± 2 2 ≈ – 0.828 or 4.828
To find the y-intercepts, let x = 0 and solve:
2
(0
+ 1)2 = 9
€ – 2) + (y
4 + (y + 1)2 = 9
(subtract 4 from both sides)
2
€
(y + 1) = 5
(use the square root property)
y+1=± 5
(subtract 1 from both sides)
5 ≈ – 3.236 or 1.236
x-intercepts: (2 – 2 2 , 0) ≈ (– 0.828, 0) & (2 + 2 2 , 0) ≈ (4.828, 0)
€
y-intercepts:
(0, – 1 – 5 ) ≈ (0, – 3.236) & (0, – 1 + 5 ) ≈ (0, 1.236)
€
€
€
€
€
y=–1±
4
Ex. 4
7x2 + 7y2 – 84x + 28y + 168 = 0
Solution:
a) First, divide both sides by seven:
x2 + y2 – 12x + 4y + 24 = 0 (rearrange terms)
x2 – 12x + y2 + 4y + 24 = 0 (subtract 24 from both sides)
x2 – 12x + y2 + 4y = – 24
Now, complete the square:
x2 – 12x +
(
−12
2
2
)
+ y2 + 4y +
2
( )
4
2
= – 24 +
(
−12
2
2
2
) ( )
+
4
2
x2 – 12x + 36 + y2 + 4y + 4 = – 24 + 36 + 4 (factor & simplify)
(x –€6)2 + (y + 2)2 = 16€
Thus, (h, k) = (6, – 2) and r = 4
€
€
b) To graph the circle, we locate
the center (6, – 2) and then plot
four points that are 4 units above,
below, to the left, and to the right
of (6, – 2). Finally, we draw a circle
through those four points (see figure
on the right).
c) To find the x-intercepts, let y = 0
and solve:
(x – 6)2 + (0 + 2)2 = 16
(x – 6)2 + 4 = 16
(subtract 4 from both sides)
2
(x – 6) = 12
(use the square root property)
x – 6 = ± 12
(add 6 to both sides and simplify)
x = 6 ± 2 3 ≈ 2.536 or 9.464
To find the y-intercepts, let x = 0 and solve:
2
2
(0
€ – 6) + (y2+ 2) = 16
36 + (y + 2) = 16
(subtract 36 from both sides)
2
€
(y + 2) = – 20
(use the square root property)
y + 2 = ± −20 which has no real solution
x-intercepts: (6 – 2 3 , 0) ≈ (2.536, 0) & (6 + 2 3 , 0) ≈ (9.464, 0)
y-intercepts: None
€
€
€
5
Find the standard form of the equation of each circle:
Ex. 5
A circle with endpoints of a diameter at (– 3, 1) and (7, 5).
Solution:
The center of the circle is the midpoint between the endpoints of the
diameter.
Midpoint =
( x +2x
1
2
,
y1 + y 2
2
,
) = ( −3+7
2
1+5
2
) = (2, 3)
Thus, (h, k) = (2, 3).
The radius of the circle is the distance from the center to one of the
points on the circle:
€
€
€
€
r = (x 2 − x1)2 +(y 2 − y1)2
(use (2, 3) and (7, 5))
=
(7−2)2 +(5−3)2
(simplify)
25+ 4 = 29
Hence, r = 29 . Plugging in (h, k) and r, the equation becomes:
€
(x – 2)2 + (y – 3)2 = 29
=
€
€
Ex. 6
€
€
The graph of the circle is given:
Solution:
The circle is centered at (– 4, 1), so (h, k) = (– 4, 1) and the distance
from the center to (– 2, 1) on the circle is 2, so r = 2. Plugging in (h, k)
and r, the equation becomes:
(x + 4)2 + (y – 1)2 = 4
Note, to graph a circle on a graphing calculator, you will need to first solve
for y. You will get two expressions, a positive square root and a negative
square root. You will then need to let Y1 = 1st expression and Y2 = 2nd
expression and graph both at the same time to get a graph of a circle.