12 multiple-choice questions, progressively harder.
What are the center and radius of the circle (x−8)2+(y+5)2=12(x - 8)^2 + (y + 5)^2 = 12(x−8)2+(y+5)2=12?
Solution
Correct answer: D
Read x−8x - 8x−8 and y+5=y−(−5)y + 5 = y - (-5)y+5=y−(−5), then simplify the square root of the right side.
h=8,k=−5,r=12=4⋅3=23h = 8, \quad k = -5, \quad r = \sqrt{12} = \sqrt{4 \cdot 3} = 2\sqrt{3}h=8,k=−5,r=12=4⋅3=23
The center is (8,−5)(8, -5)(8,−5) and the radius is 232\sqrt{3}23.
Which equation describes the circle with center (−2,7)(-2, 7)(−2,7) that passes through (−2,1)(-2, 1)(−2,1)?
Correct answer: A
The radius is the distance from (−2,7)(-2, 7)(−2,7) to (−2,1)(-2, 1)(−2,1), a purely vertical gap, so r=6r = 6r=6 and r2=36r^2 = 36r2=36.
(x+2)2+(y−7)2=62=36(x + 2)^2 + (y - 7)^2 = 6^2 = 36(x+2)2+(y−7)2=62=36
The center −2-2−2 makes x+2x + 2x+2, and the right side is 363636.
Describe the graph of x2+y2+2x+2y+2=0x^2 + y^2 + 2x + 2y + 2 = 0x2+y2+2x+2y+2=0.
Correct answer: C
Complete the square: half of 222 is 111 (square 111) for each variable.
(x+1)2+(y+1)2=−2+1+1=0(x + 1)^2 + (y + 1)^2 = -2 + 1 + 1 = 0(x+1)2+(y+1)2=−2+1+1=0
A sum of two squares is 000 only when both are 000, so the graph is the single point (−1,−1)(-1, -1)(−1,−1).
Describe the graph of x2+y2−4x+10=0x^2 + y^2 - 4x + 10 = 0x2+y2−4x+10=0.
There is no yyy linear term. Complete the square on x2−4xx^2 - 4xx2−4x: half of −4-4−4 is −2-2−2 (square 444).
(x−2)2+y2=−10+4=−6(x - 2)^2 + y^2 = -10 + 4 = -6(x−2)2+y2=−10+4=−6
A sum of two squares cannot be negative, so no real point satisfies it and there is no graph.
Write the equation of the circle with center (0,3)(0, 3)(0,3) and radius 222 in general form.
Start in standard form and expand: x2+(y−3)2=4x^2 + (y - 3)^2 = 4x2+(y−3)2=4.
x2+y2−6y+9=4⇒x2+y2−6y+5=0x^2 + y^2 - 6y + 9 = 4 \Rightarrow x^2 + y^2 - 6y + 5 = 0x2+y2−6y+9=4⇒x2+y2−6y+5=0
Moving the 444 across leaves the constant 9−4=59 - 4 = 59−4=5.
What is the radius of the circle x2+y2−14x+48=0x^2 + y^2 - 14x + 48 = 0x2+y2−14x+48=0?
Correct answer: B
There is no yyy linear term. Complete the square on x2−14xx^2 - 14xx2−14x: half of −14-14−14 is −7-7−7 (square 494949).
(x−7)2+y2=−48+49=1(x - 7)^2 + y^2 = -48 + 49 = 1(x−7)2+y2=−48+49=1
The radius is 1=1\sqrt{1} = 11=1.
Which point lies on the circle x2+y2=50x^2 + y^2 = 50x2+y2=50?
A point is on the circle when x2+y2=50x^2 + y^2 = 50x2+y2=50. Test each candidate.
52+52=25+25=50✓5^2 + 5^2 = 25 + 25 = 50 \checkmark52+52=25+25=50✓
Only (5,5)(5, 5)(5,5) works; (6,4)(6, 4)(6,4) gives 525252, (3,6)(3, 6)(3,6) gives 454545, and (0,7)(0, 7)(0,7) gives 494949.
What is the center of the circle 3x2+3y2+6x−18y+3=03x^2 + 3y^2 + 6x - 18y + 3 = 03x2+3y2+6x−18y+3=0?
The x2x^2x2 and y2y^2y2 coefficients are both 333, so divide through by 333 first: x2+y2+2x−6y+1=0x^2 + y^2 + 2x - 6y + 1 = 0x2+y2+2x−6y+1=0. Then complete the square.
(x+1)2+(y−3)2=−1+1+9=9(x + 1)^2 + (y - 3)^2 = -1 + 1 + 9 = 9(x+1)2+(y−3)2=−1+1+9=9
The center is (−1,3)(-1, 3)(−1,3).
Where does the circle (x−3)2+(y−4)2=25(x - 3)^2 + (y - 4)^2 = 25(x−3)2+(y−4)2=25 cross the xxx-axis?
On the xxx-axis, y=0y = 0y=0. Substitute and solve for xxx.
(x−3)2+(0−4)2=25⇒(x−3)2=9⇒x−3=±3(x - 3)^2 + (0 - 4)^2 = 25 \Rightarrow (x - 3)^2 = 9 \Rightarrow x - 3 = \pm 3(x−3)2+(0−4)2=25⇒(x−3)2=9⇒x−3=±3
So x=6x = 6x=6 or x=0x = 0x=0. Among the options, (6,0)(6, 0)(6,0) is a crossing point.
Which equation describes the circle with center (−1,−1)(-1, -1)(−1,−1) that passes through (2,3)(2, 3)(2,3)?
The radius squared is the distance squared from (−1,−1)(-1, -1)(−1,−1) to (2,3)(2, 3)(2,3).
r2=(2−(−1))2+(3−(−1))2=32+42=25r^2 = (2 - (-1))^2 + (3 - (-1))^2 = 3^2 + 4^2 = 25r2=(2−(−1))2+(3−(−1))2=32+42=25
With center (−1,−1)(-1, -1)(−1,−1), the equation is (x+1)2+(y+1)2=25(x + 1)^2 + (y + 1)^2 = 25(x+1)2+(y+1)2=25.
The center of the circle x2+y2+Dx+Ey+F=0x^2 + y^2 + Dx + Ey + F = 0x2+y2+Dx+Ey+F=0 is (−3,5)(-3, 5)(−3,5). What are DDD and EEE?
The general-form center is (−D2,−E2)\left(-\tfrac{D}{2}, -\tfrac{E}{2}\right)(−2D,−2E). Set each equal to the given center.
−D2=−3⇒D=6,−E2=5⇒E=−10-\tfrac{D}{2} = -3 \Rightarrow D = 6, \qquad -\tfrac{E}{2} = 5 \Rightarrow E = -10−2D=−3⇒D=6,−2E=5⇒E=−10
So D=6D = 6D=6 and E=−10E = -10E=−10.
The circle x2+y2=25x^2 + y^2 = 25x2+y2=25 meets the negative xxx-axis at which point?
On the xxx-axis, y=0y = 0y=0, so x2=25x^2 = 25x2=25 and x=±5x = \pm 5x=±5.
x2=25⇒x=±5x^2 = 25 \Rightarrow x = \pm 5x2=25⇒x=±5
The negative xxx-axis crossing is (−5,0)(-5, 0)(−5,0); the points (0,±5)(0, \pm 5)(0,±5) are on the yyy-axis.
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