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 Squaring
both sides, we obtain the standard equation: (x — a)2 + (y — b)2 = r2.
4.2 Write the standard equation for the circle with center (3,5) and radius 4.
(x-3)2 + (y-5)2 = 16.
4.3 Write the standard equation for the circle with center (4, -2) and radius 7.
(;t-4)2 + (>>+2)2 = 49.
4.4 Write the standard equation for the circle with center at the origin and radius r.
X2 + y2 = r2.
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.
4.6 Identify the graph of the equation x2 + y2 - 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.
4.7 Identify the graph of the equation x2 + y2 + 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.
4.8 Identify the graph of the equation x2 + y2 + 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.
4.9 Show that any circle has an equation of the form x2 + y2 + Dx + Ey + F = 0.
Consider the standard equation (x - a)2 + (y - b)2 = r2. Squaring and simplifying, x2 + y2 — lax —
2by + a2 + b2-r2 = 0. Let D = -2a, E = -2b, and F = a2 + b2 - r2.
4.10 Determine the graph of an equation x2 + y2 + Dx + Ey + F = 0.
I Complete the square: Simplifying:
Now, let d = D2+E2. When d>0. we obtain a circle with center
at and radius When d=0, we obtain a single point when d<0,
graph contains no points at all.
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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 Squaring
both sides, we obtain the standard equation: (x — a)2 + (y — b)2 = r2.
4.2 Write the standard equation for the circle with center (3,5) and radius 4.
(x-3)2 + (y-5)2 = 16.
4.3 Write the standard equation for the circle with center (4, -2) and radius 7.
(;t-4)2 + (>>+2)2 = 49.
4.4 Write the standard equation for the circle with center at the origin and radius r.
X2 + y2 = r2.
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.
4.6 Identify the graph of the equation x2 + y2 - 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.
4.7 Identify the graph of the equation x2 + y2 + 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.
4.8 Identify the graph of the equation x2 + y2 + 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.
4.9 Show that any circle has an equation of the form x2 + y2 + Dx + Ey + F = 0.
Consider the standard equation (x - a)2 + (y - b)2 = r2. Squaring and simplifying, x2 + y2 — lax —
2by + a2 + b2-r2 = 0. Let D = -2a, E = -2b, and F = a2 + b2 - r2.
4.10 Determine the graph of an equation x2 + y2 + Dx + Ey + F = 0.
I Complete the square: Simplifying:
Now, let d = D2+E2. When d>0. we obtain a circle with center
at and radius When d=0, we obtain a single point when d<0,
graph contains no points at all.
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