CHAPTER 4
Circles
4.1
Write the standard equation for a circle with center at (a, b) and radius r.
y the distance formula, a point (x, y) is on the circle if and only if
both sides, we obtain the standard equation: (x — a)
2 + (y — b)
2 = r
2 .
4.2
Write the standard equation for the circle with center (3,5) and radius 4.
4.3
Write the standard equation for the circle with center (4, -2) and radius 7.
4.4
Write the standard equation for the circle with center at the origin and radius r.
4.5
Find the standard equation of the circle with center at (1, -2) and passing through the point (7, 4).
The radius of the circle is the distance between (1, -2) and (7, 4):
V72. Thus, the standard equation is: (x - I)
2 + (y + 2)
2 = 72.
Identify the graph of the equation x
2 + y
2 - I2x + 20y + 15 = 0.
Complete the square in x and in y: (x - 6)
2 + (y + 10)
2 + 15 = 36 + 100. [Here the-6 in (x - 6) is half
of the coefficient, -12, of x in the original equation, and the + 10in (_y + 10) is half of the coefficient 20, of y.
The 36 and 100 on the right balance the squares of -6 and +10 that have in effect been added on the left.]
Simplifying, we obtain (x - 6)
2 + (y + 10)
2 = 121, the standard equation of a circle with center at (6, -10) and
radius 11.
Identify the graph of the equation x
2 + y
2 + 3x — 2y + 4 = 0.
Complete the square (as in Problem 4.6): (jc + |)
2 + (y - I)
2 + 4 = j + 1. Simplifying, we obtain
(x + 1 )
2 + (y - I)
2 = ~ 1. But this equation has no solutions, since the left side is always nonnegative. In
other words, the graph is the empty set.
Identify the graph of the equation x
2 + y
2 + 2x - 2y + 2 = 0.
Complete the square: (x + I)
2 + (y - I)
2 + 2 = 1 + 1, which simplifies to (x + I)
2 + (y - I)
2 = 0. This
is satisfied when and only when * + l=0 and y —1=0, that is, for the point (—1,1). Hence, the graph is
a single point.
Show that any circle has an equation of the form x
2 + y
2 + Dx + Ey + F = 0.
Consider the standard equation (x - a)
2 + (y - b)
2 = r
2 . Squaring and simplifying, x
2 + y
2 — lax —
2by + a
2 + b
2 -r
2 = 0. Let D = -2a, E = -2b, and F = a
2 + b
2 - r
2 .
Determine the graph of an equation x
2 + y
2 + Dx + Ey + F = 0.
at
graph contains no points at all.
19
(x-3)2 + (y-5)2 = 16.
(;t-4)2 + (>>+2)2 = 49.
X + y2 = r2.
2
4.6
4.7
4.8
4.9
4.10
I Complete the square:
Simplifying:
and radius
When
Now, let
When
we obtain a circle with center
Squaring
when d<0,
d = D2+E2.
d >0.
we obtain a single point
d=0,
Circles
4.1
Write the standard equation for a circle with center at (a, b) and radius r.
y the distance formula, a point (x, y) is on the circle if and only if
both sides, we obtain the standard equation: (x — a)
2 + (y — b)
2 = r
2 .
4.2
Write the standard equation for the circle with center (3,5) and radius 4.
4.3
Write the standard equation for the circle with center (4, -2) and radius 7.
4.4
Write the standard equation for the circle with center at the origin and radius r.
4.5
Find the standard equation of the circle with center at (1, -2) and passing through the point (7, 4).
The radius of the circle is the distance between (1, -2) and (7, 4):
V72. Thus, the standard equation is: (x - I)
2 + (y + 2)
2 = 72.
Identify the graph of the equation x
2 + y
2 - I2x + 20y + 15 = 0.
Complete the square in x and in y: (x - 6)
2 + (y + 10)
2 + 15 = 36 + 100. [Here the-6 in (x - 6) is half
of the coefficient, -12, of x in the original equation, and the + 10in (_y + 10) is half of the coefficient 20, of y.
The 36 and 100 on the right balance the squares of -6 and +10 that have in effect been added on the left.]
Simplifying, we obtain (x - 6)
2 + (y + 10)
2 = 121, the standard equation of a circle with center at (6, -10) and
radius 11.
Identify the graph of the equation x
2 + y
2 + 3x — 2y + 4 = 0.
Complete the square (as in Problem 4.6): (jc + |)
2 + (y - I)
2 + 4 = j + 1. Simplifying, we obtain
(x + 1 )
2 + (y - I)
2 = ~ 1. But this equation has no solutions, since the left side is always nonnegative. In
other words, the graph is the empty set.
Identify the graph of the equation x
2 + y
2 + 2x - 2y + 2 = 0.
Complete the square: (x + I)
2 + (y - I)
2 + 2 = 1 + 1, which simplifies to (x + I)
2 + (y - I)
2 = 0. This
is satisfied when and only when * + l=0 and y —1=0, that is, for the point (—1,1). Hence, the graph is
a single point.
Show that any circle has an equation of the form x
2 + y
2 + Dx + Ey + F = 0.
Consider the standard equation (x - a)
2 + (y - b)
2 = r
2 . Squaring and simplifying, x
2 + y
2 — lax —
2by + a
2 + b
2 -r
2 = 0. Let D = -2a, E = -2b, and F = a
2 + b
2 - r
2 .
Determine the graph of an equation x
2 + y
2 + Dx + Ey + F = 0.
at
graph contains no points at all.
19
(x-3)2 + (y-5)2 = 16.
(;t-4)2 + (>>+2)2 = 49.
X + y2 = r2.
2
4.6
4.7
4.8
4.9
4.10
I Complete the square:
Simplifying:
and radius
When
Now, let
When
we obtain a circle with center
Squaring
when d<0,
d = D2+E2.
d >0.
we obtain a single point
d=0,