12 multiple-choice questions, progressively harder.
What is the radius of the circle (x−2)2+(y−6)2=20(x - 2)^2 + (y - 6)^2 = 20(x−2)2+(y−6)2=20?
Solution
Correct answer: D
The radius is the square root of the right side. Simplify 20\sqrt{20}20 by pulling out the perfect-square factor.
r=20=4⋅5=25r = \sqrt{20} = \sqrt{4 \cdot 5} = 2\sqrt{5}r=20=4⋅5=25
The radius is 252\sqrt{5}25, not 202020.
What is the distance between the points (1,2)(1, 2)(1,2) and (4,6)(4, 6)(4,6)?
Correct answer: A
Use the distance formula with horizontal gap 4−1=34 - 1 = 34−1=3 and vertical gap 6−2=46 - 2 = 46−2=4.
d=32+42=9+16=25=5d = \sqrt{3^2 + 4^2} = \sqrt{9 + 16} = \sqrt{25} = 5d=32+42=9+16=25=5
The points are 555 apart.
Which equation describes the circle with center (0,−4)(0, -4)(0,−4) and radius 222?
With h=0h = 0h=0 the xxx-term is just x2x^2x2, and k=−4k = -4k=−4 gives y−(−4)=y+4y - (-4) = y + 4y−(−4)=y+4; the right side is r2r^2r2.
x2+(y+4)2=22=4x^2 + (y + 4)^2 = 2^2 = 4x2+(y+4)2=22=4
The center below the origin makes y+4y + 4y+4, and the right side is 444.
A circle has center (1,1)(1, 1)(1,1) and passes through (1,6)(1, 6)(1,6). What is its radius?
Correct answer: B
The radius is the distance from (1,1)(1, 1)(1,1) to (1,6)(1, 6)(1,6). The points share an xxx-coordinate, so this is just the vertical gap.
r=(1−1)2+(6−1)2=0+25=5r = \sqrt{(1 - 1)^2 + (6 - 1)^2} = \sqrt{0 + 25} = 5r=(1−1)2+(6−1)2=0+25=5
The radius is 555.
What is the center of the circle (x−12)2+(y+32)2=4\left(x - \tfrac{1}{2}\right)^2 + \left(y + \tfrac{3}{2}\right)^2 = 4(x−21)2+(y+23)2=4?
Correct answer: C
Read x−12x - \tfrac{1}{2}x−21 directly and rewrite y+32y + \tfrac{3}{2}y+23 as y−(−32)y - \left(-\tfrac{3}{2}\right)y−(−23).
h=12,k=−32h = \tfrac{1}{2}, \qquad k = -\tfrac{3}{2}h=21,k=−23
The center is (12,−32)\left(\tfrac{1}{2}, -\tfrac{3}{2}\right)(21,−23).
What is the distance between the origin and the point (−5,−12)(-5, -12)(−5,−12)?
Square each coordinate, add, then take the root.
d=(−5)2+(−12)2=25+144=169=13d = \sqrt{(-5)^2 + (-12)^2} = \sqrt{25 + 144} = \sqrt{169} = 13d=(−5)2+(−12)2=25+144=169=13
The distance is 131313.
Which point does NOT lie on the circle x2+y2=169x^2 + y^2 = 169x2+y2=169?
Test each point in x2+y2=169x^2 + y^2 = 169x2+y2=169. Three of them satisfy it.
52+122=169,132+02=169,(−12)2+52=169,102+102=2005^2 + 12^2 = 169, \quad 13^2 + 0^2 = 169, \quad (-12)^2 + 5^2 = 169, \quad 10^2 + 10^2 = 20052+122=169,132+02=169,(−12)2+52=169,102+102=200
Only (10,10)(10, 10)(10,10) gives 200≠169200 \ne 169200=169, so it is not on the circle.
Which equation describes the circle with center (−3,−3)(-3, -3)(−3,−3) and radius 333?
Both center coordinates are −3-3−3, so both parentheses become +3+3+3; the right side is r2r^2r2.
(x−(−3))2+(y−(−3))2=32⇒(x+3)2+(y+3)2=9(x - (-3))^2 + (y - (-3))^2 = 3^2 \Rightarrow (x + 3)^2 + (y + 3)^2 = 9(x−(−3))2+(y−(−3))2=32⇒(x+3)2+(y+3)2=9
The right side is 999, not 333.
The point (6,0)(6, 0)(6,0) lies on a circle centered at the origin. What is the circle's equation?
The radius is the distance from (0,0)(0, 0)(0,0) to (6,0)(6, 0)(6,0), which is 666, so r2=36r^2 = 36r2=36.
x2+y2=62=36x^2 + y^2 = 6^2 = 36x2+y2=62=36
The circle is centered at the origin, so the equation is x2+y2=36x^2 + y^2 = 36x2+y2=36.
What are the center and radius of the circle (x+10)2+(y−4)2=100(x + 10)^2 + (y - 4)^2 = 100(x+10)2+(y−4)2=100?
Rewrite x+10x + 10x+10 as x−(−10)x - (-10)x−(−10) and read y−4y - 4y−4; the radius is the square root of the right side.
h=−10,k=4,r=100=10h = -10, \quad k = 4, \quad r = \sqrt{100} = 10h=−10,k=4,r=100=10
The center is (−10,4)(-10, 4)(−10,4) and the radius is 101010, not 100100100.
What is the distance between the points (2,2)(2, 2)(2,2) and (2,−3)(2, -3)(2,−3)?
The points share an xxx-coordinate, so the distance is the size of the vertical gap.
d=(2−2)2+(−3−2)2=0+25=5d = \sqrt{(2 - 2)^2 + (-3 - 2)^2} = \sqrt{0 + 25} = 5d=(2−2)2+(−3−2)2=0+25=5
The distance is 555.
According to the distance formula, the distance between the point (a,b)(a, b)(a,b) and the origin is:
Apply the distance formula with the second point at the origin (0,0)(0, 0)(0,0).
d=(a−0)2+(b−0)2=a2+b2d = \sqrt{(a - 0)^2 + (b - 0)^2} = \sqrt{a^2 + b^2}d=(a−0)2+(b−0)2=a2+b2
The distance squares each coordinate, adds, then takes the root, so it is a2+b2\sqrt{a^2 + b^2}a2+b2.
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