MATH 150
Quiz Key #3
9/29-10/1/2015
(1) Write the equation of a circle with center C (−2, 4) and radius 5 in standard
form.
Solution: Standard form for the equation of a circle is ( x − h)2 + (y − k )2 = r2 ,
where (h, k) is the center and r is the radius. Hence, the required equation is:
( x + 2)2 + (y − 4)2 = 25.
(2) Find the distance between A (−5, −4) and B (6, 3) , and then find the midpoint of the line segment that connects them.
q
Solution: The distance formula is d = ( x2 − x1 )2 + (y2 − y1 )2 . Therefore, the
distance between A and B is given by the following:
q
p
d ( A, B) = (6 − (−5))2 + (3 − (−4))2 = 112 + 72
=
√
121 + 49 =
√
170.
To check whether the radical can be simplified,
√ we need to check the prime
factorization of 170. 170 = 2· 5 · 17. Therefore,
170 is simplest form.
The midpoint formula is
x1 + x2 y1 + y2
2 ,
2
. Hence, the midpoint of AB is given
by
−5 + 6 −4 + 3
,
2
2
=
1 1
,−
2 2
.
1
(3) Convert the equation to standard form of a circle. Then find the center,
radius, domain, and range.
3x2 + 3y2 − 5x + 6y − 4 = 0.
(Standard Form)
1
1 2
3x + 3y2 − 5x + 6y − 4 = · 0.
3
3
5
4
x2 + y2 − x + 2y − = 0.
3
3
5
4
x2 − x + y2 + 2y = .
3
3
2 ! 2
5
4
5
5
2
2
+ y + 2y + 1 = +
+ 1.
x − x+
3
6
3
6
Note that
4
+
3
2
5
48 25 36
109
+1 =
+
+
=
.
6
36 36 36
36
Since the only prime factors of 36 are 2 and 3, neither of which are factors of 109,
this fraction is in simplest form. Therefore, the standard form of the equation
in the problem is
5 2
109
x−
+ ( y + 1)2 =
.
6
36
The center is 65 , −1 .
√
The radius is h109
6 . √
i
√
109
5
The domain is 56 − 109
,
+
.
6
6
6
i
h
√
√
109
.
The range is −1 − 109
6 , −1 + 6
2
(4) If the distance between A (2, y) and B (7, 1) is 13 units, find all possible
values for y.
From the distance formula, we know that any possible values of y have to be
solutions to the following equation:
q
(2 − 7)2 + (y − 1)2 = 13.
Therefore, if we solve this equation, that should give the possibly values for y.
Squaring both sides of the equation yields
(2 − 7)2 + (y − 1)2 = 169.
25 + (y − 1)2 = 169.
√
(y − 1)2 = 144 =⇒ y − 1 = ± 144 = ±12.
Therefore, y = 1 ± 12. Thus, y = 13 or y = −11.
(5) If (−2, 4) and (2, 0) are endpoints of a diameter of a circle, write the equation for the circle in standard form.
Solution: To write the equation of a circle in standard form, we need to know
its center and radius. The center is the midpoint of the diameter. Therefore, by
using the midpoint formula, we know that the center of the circle is
−2 + 2 4 + 0
,
= (0, 2) .
2
2
Now the radius of the circle is the distance from its center to any point on the
circle. Therefore, the radius is the distance between (0, 2) and (−2, 4) , and is
also the distance between (0, 2) and (2, 0) . I will find the second distance (since
the two distances are the same, it doesn’t matter which one).
q
√
√
√
r = (0 − 2)2 + (2 − 0)2 = 4 + 4 = 8 = 2 2.
Therefore, standard form of the circle is
√ 2
( x − 0)2 + ( y − 2)2 = 2 2 ,
which simplifies to
x2 + (y − 2)2 = 8.
3
(6) Convert the following angles from degrees to radians:
180◦
60◦
330◦
240◦
− 45◦
Solution: There are 2π radians in a circle and 360◦ . Therefore, for every 180◦
there are π radians. Using this conversion factor, we can find the following:
180◦ = π
240◦ =
60◦ =
π
60π
=
180
3
240π
4π
=
180
3
330◦ =
− 45◦ =
330π
11π
=
180
6
−45π
π
=− .
180
4
(7) Label ALL of the following angles on the unit circle in radians.
Degrees
0◦
30◦
45◦
60◦
90◦
120◦
135◦
150◦
180◦
210◦
225◦
240◦
270◦
300◦
315◦
330◦
Radians
0, 2π
π
6
π
4
π
3
π
2
2π
3
3π
4
5π
6
π
7π
6
5π
4
4π
3
3π
2
5π
3
7π
4
11π
6
4