Chapter Twenty-Two — Conic Sections
The Humongous Book of Algebra Problems
494
Apply cross multiplication to solve the proportion.
Substitute , , and k = 0 into the appropriate standard form
equation.
Circles
Center, radius, and diameter
Note: Problems 22.12–22.13 refer to the circle with center (2,–1) and radius 4.
22.12 Graph the circle.
Plot the center point (2,–1) on the coordinate plane. Next, plot the points that
are four units above, below, left, and right of the center. Draw the circle that
passes through those four points, as illustrated by Figure 22-3.
Figure 22-3: Each of the points on the graph of the circle is exactly four units away from
the center point (2,–1).
Use this one
instead of
y = a(x – h)
2
+ k
because the directrix
is vertical and the
axis of symmetry is
horizontal.
The points are
(2,3), (2,–5),
(–2,–1), and (6,–1).
Chapter Twenty-Two — Conic Sections
The Humongous Book of Algebra Problems
495
Note: Problems 22.12–22.13 refer to the circle with center (2,–1) and radius 4.
22.13 Write the equation of the circle in standard form.
The standard form for the equation of a circle is (x – h)
2
+ (y – k)
2
= r
2
, where
(h,k) is the center of the circle and r is the radius. Substitute h = 2, k = –1, and
r = 4 into the equation.
Note: Problems 22.14–22.16 refer to the equation x
2
+ y
2
+ 8x – 6y = 0.
22.14 Write the equation of the circle in standard form.
Use parentheses to group the x-terms and the y-terms.
(x
2
+ 8x) + (y
2
– 6y) = 0
Complete the square twice, once for each parenthetical quantity.
Note: Problems 22.14–22.16 refer to the equation x
2
+ y
2
+ 8x – 6y = 0.
22.15 Identify the center and radius of the circle.
The standard form of a circle is (x – h)
2
+ (y – k)
2
= r
2
. According to Problem
22.14, the standard form of this circle is (x + 4)
2
+ (y – 3)
2
= 25. Therefore,
h = –4, k = 3, and r
2
= 25. Solve the equation to calculate r.
The solution r = –5 is discarded, as the radius of a circle must be a positive
number. The center of the circle is (h,k) = (–4,3) and the radius is r = 5.
Square
half of the
x-coefcient:
(8 ÷ 2)
2
= 4
2
= 16.
Add that number
inside the left set
of parentheses and
make sure to add
it to the right side
of the equation, too.
Then, square half of
the y-coefcient:
(–6 ÷ 2)
2
= (–3)
2
= 9.
Add that to the
second set of
parentheses and
the right side of
the equation.
h is the opposite
of the number in the
x parentheses and k
is the opposite of the
number in the y
parentheses.
Chapter Twenty-Two — Conic Sections
The Humongous Book of Algebra Problems
496
Note: Problems 22.14–22.16 refer to the equation x
2
+ y
2
+ 8x – 6y = 0.
22.16 Graph the circle.
According to Problem 22.15, the center of the circle is (–4,3) and the radius
is 5. Plot the center and then plot the points five units above, below, left, and
right of the center. Draw a circle that intersects the four points surrounding the
center, as illustrated in Figure 22-4.
Figure 22-4: The graph of the circle x
2
+ y
2
+ 8x – 6y = 0 has center (–4,3) and
radius 5.
22.17 Write the equation of the circle 3x
2
+ 3y
2
+ 12x + 15y – 1 = 0 in standard form
and identify the center and radius.
Writing the equation in standard form requires you to complete the square.
Divide each of the terms in the equation by 3 so that the coefficients of x
2
and
y
2
are 1.
Move the constant to the right side of the equation by adding to both sides.
The x
2
- and
y
2
-terms have the
same coefcient in the
equations of circles (in
this case 3).
Chapter Twenty-Two — Conic Sections
The Humongous Book of Algebra Problems
497
Place the x- and y-terms in separate groups.
Complete the square twice, once for each group of terms on the left side of the
equation.
The equation is in standard form (x – h)
2
+ (y – k)
2
= r
2
so the center of the circle
is . Calculate the radius.
Note: Problems 22.18–22.19 refer to the circle with center (–6,4) that passes through
point (–2,3).
22.18 Use the distance formula to calculate the radius of
the circle.
The distance between the center of a circle and any point on that circle is
equal to the radius. Substitute x
1
= –6, y
1
= 4, x
2
= –2, and y
2
= 3 into the distance
formula to calculate the radius.
To rationalize
the denominator,
multiply the
numerator and
denominator by
.
x
1
and y
1
refer to the
x- and y-values
of point #1, which
could be (–6,4) OR
(–2,3). In other words,
substituting x
1
= –2,
y
1
= 3, x
2
= –6, and
y
2
= 4 into the
distance formula
gives you the
same nal
answer.
Chapter Twenty-Two — Conic Sections
The Humongous Book of Algebra Problems
498
Note: Problems 22.18–22.19 refer to the circle with center (–6,4) that passes through
point (–2,3).
22.19 Write the equation of the circle in standard form.
The center of the circle is (–6,4), so h = –6 and k = 4. According to Problem
22.18, the radius of the circle is . Substitute h, k, and r into the standard
form equation.
Note: Problems 22.20–22.21 refer to the circle that has a diameter with endpoints (3,–1)
and (–3,7).
22.20 Use the midpoint formula to identify the center of the circle.
The midpoint of a diameter is the center of the corresponding circle. Apply the
midpoint formula to the endpoints of the diameter to identify the center (h,k)
of the circle.
Note: Problems 22.20–22.21 refer to the circle that has a diameter with endpoints (3,–1)
and (–3,7).
22.21 Calculate the radius of the circle and write the equation in standard form.
Apply the distance formula, defined in Problem 22.18, to calculate the radius.
The x-value
of the midpoint is
the average of the
endpoints’ x-values,
so add them up and
divide by 2. Do the
same thing with
the y’s to complete
the midpoint.
Find the
distance between
the center (0,3) and
either of the endpoints.
For example, you could
set x
1
= 0 and y
1
= 3
(the center) and set
x
2
= 3 and y
2
= –1 (the
rst of the endpoints).
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