12 multiple-choice questions, progressively harder.
What is the center of the circle (x+3)2+(y−2)2=16(x + 3)^2 + (y - 2)^2 = 16(x+3)2+(y−2)2=16?
Solution
Correct answer: B
The form subtracts the center coordinate, so rewrite x+3x + 3x+3 as x−(−3)x - (-3)x−(−3).
x+3=x−(−3)⇒h=−3,y−2⇒k=2x + 3 = x - (-3) \Rightarrow h = -3, \qquad y - 2 \Rightarrow k = 2x+3=x−(−3)⇒h=−3,y−2⇒k=2
The center is (−3,2)(-3, 2)(−3,2).
Which equation describes the circle centered at the origin with radius 666?
A circle centered at the origin has equation x2+y2=r2x^2 + y^2 = r^2x2+y2=r2, and the right side is the radius squared.
r=6⇒r2=36⇒x2+y2=36r = 6 \Rightarrow r^2 = 36 \Rightarrow x^2 + y^2 = 36r=6⇒r2=36⇒x2+y2=36
The right side must be 62=366^2 = 3662=36, not 666.
Which equation describes the circle with center (4,1)(4, 1)(4,1) and radius 222?
Correct answer: D
Substitute h=4h = 4h=4, k=1k = 1k=1, and r=2r = 2r=2 into (x−h)2+(y−k)2=r2(x - h)^2 + (y - k)^2 = r^2(x−h)2+(y−k)2=r2.
(x−4)2+(y−1)2=22=4(x - 4)^2 + (y - 1)^2 = 2^2 = 4(x−4)2+(y−1)2=22=4
The center coordinates are subtracted, and the right side is r2=4r^2 = 4r2=4.
What is the radius of the circle (x−7)2+(y+3)2=81(x - 7)^2 + (y + 3)^2 = 81(x−7)2+(y+3)2=81?
Correct answer: A
The right side is the radius squared, so take its square root.
r=81=9r = \sqrt{81} = 9r=81=9
The radius is 999.
What is the center of the circle (x−6)2+(y+8)2=1(x - 6)^2 + (y + 8)^2 = 1(x−6)2+(y+8)2=1?
Correct answer: C
Read each parenthesis against (x−h)2+(y−k)2(x - h)^2 + (y - k)^2(x−h)2+(y−k)2, remembering that y+8=y−(−8)y + 8 = y - (-8)y+8=y−(−8).
x−6⇒h=6,y+8⇒k=−8x - 6 \Rightarrow h = 6, \qquad y + 8 \Rightarrow k = -8x−6⇒h=6,y+8⇒k=−8
The center is (6,−8)(6, -8)(6,−8).
What is the center of the circle (x−9)2+y2=4(x - 9)^2 + y^2 = 4(x−9)2+y2=4?
The term y2y^2y2 is (y−0)2(y - 0)^2(y−0)2, so the yyy-coordinate of the center is 000.
x−9⇒h=9,y2=(y−0)2⇒k=0x - 9 \Rightarrow h = 9, \qquad y^2 = (y - 0)^2 \Rightarrow k = 0x−9⇒h=9,y2=(y−0)2⇒k=0
The center is (9,0)(9, 0)(9,0).
Which equation describes the circle centered at the origin with radius 111 (the unit circle)?
Use x2+y2=r2x^2 + y^2 = r^2x2+y2=r2 with r=1r = 1r=1.
x2+y2=12=1x^2 + y^2 = 1^2 = 1x2+y2=12=1
The unit circle is x2+y2=1x^2 + y^2 = 1x2+y2=1.
What is the center of the circle (x+5)2+(y−7)2=36(x + 5)^2 + (y - 7)^2 = 36(x+5)2+(y−7)2=36?
Rewrite x+5x + 5x+5 as x−(−5)x - (-5)x−(−5) and read y−7y - 7y−7 directly.
x+5⇒h=−5,y−7⇒k=7x + 5 \Rightarrow h = -5, \qquad y - 7 \Rightarrow k = 7x+5⇒h=−5,y−7⇒k=7
The center is (−5,7)(-5, 7)(−5,7).
What is the radius of the circle (x−3)2+(y−3)2=2(x - 3)^2 + (y - 3)^2 = 2(x−3)2+(y−3)2=2?
The right side is r2=2r^2 = 2r2=2, which is not a perfect square, so leave the radius as a root.
r=2r = \sqrt{2}r=2
The radius is 2\sqrt{2}2, not 222.
Which equation describes the circle with center (1,−6)(1, -6)(1,−6) and radius 101010?
Substitute h=1h = 1h=1, k=−6k = -6k=−6, r=10r = 10r=10, noting y−(−6)=y+6y - (-6) = y + 6y−(−6)=y+6.
(x−1)2+(y+6)2=102=100(x - 1)^2 + (y + 6)^2 = 10^2 = 100(x−1)2+(y+6)2=102=100
The right side is r2=100r^2 = 100r2=100, not 101010.
A circle is centered at the origin and passes through (0,5)(0, 5)(0,5). What is its radius?
The radius is the distance from the center (0,0)(0, 0)(0,0) to the point (0,5)(0, 5)(0,5).
r=(0−0)2+(5−0)2=25=5r = \sqrt{(0 - 0)^2 + (5 - 0)^2} = \sqrt{25} = 5r=(0−0)2+(5−0)2=25=5
The radius is 555.
What is the center of the circle (x−2)2+(y+9)2=5(x - 2)^2 + (y + 9)^2 = 5(x−2)2+(y+9)2=5?
Read x−2x - 2x−2 directly and rewrite y+9y + 9y+9 as y−(−9)y - (-9)y−(−9).
x−2⇒h=2,y+9⇒k=−9x - 2 \Rightarrow h = 2, \qquad y + 9 \Rightarrow k = -9x−2⇒h=2,y+9⇒k=−9
The center is (2,−9)(2, -9)(2,−9).
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