Math 2 Quiz: Completing Square For Circle Equations
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Completing Square For Circle EquationsQuestion 1 of 14

The equation x2+y2−6x+10y+k=0x^2 + y^2 - 6x + 10y + k = 0 represents a circle with radius 5. What is the value of kk?

2525
−9-9
99
−25-25
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Math 2 Quiz

Math 2 Quiz: Completing Square For Circle Equations

Practice Completing Square For Circle Equations in Math 2 with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

What this quiz covers

This quiz focuses on Completing Square For Circle Equations, giving you a quick way to practice the rules, question types, and explanations that matter most for Math 2.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

All questions

Question 1

The equation x2+y2−6x+10y+k=0x^2 + y^2 - 6x + 10y + k = 0 represents a circle with radius 5. What is the value of kk?

  1. 2525
  2. −9-9
  3. 99 (correct answer)
  4. −25-25
Explanation: When you see a circle equation in general form like this, you need to convert it to standard form to identify the center and radius. This requires completing the square for both variables. Starting with x2+y2−6x+10y+k=0x^2 + y^2 - 6x + 10y + k = 0, rearrange and group the xx and yy terms: (x2−6x)+(y2+10y)=−k(x^2 - 6x) + (y^2 + 10y) = -k. To complete the square for x2−6xx^2 - 6x: take half of the coefficient of xx (which is −6-6), square it: (−3)2=9(-3)^2 = 9. So x2−6x+9=(x−3)2x^2 - 6x + 9 = (x-3)^2. For y2+10yy^2 + 10y: take half of 1010, square it: (5)2=25(5)^2 = 25. So y2+10y+25=(y+5)2y^2 + 10y + 25 = (y+5)^2. Adding these completed squares to both sides: (x−3)2+(y+5)2=−k+9+25=34−k(x-3)^2 + (y+5)^2 = -k + 9 + 25 = 34 - k. Since the radius is 5, we know r2=25r^2 = 25. The standard form is (x−3)2+(y+5)2=25(x-3)^2 + (y+5)^2 = 25, so 34−k=2534 - k = 25. Therefore, k=9k = 9. Looking at the wrong answers: (A) 25 gives you the radius squared, but that's not what kk represents in the original equation. (B) -9 would result if you forgot to add the completed square constants to the right side. (D) -25 might come from confusing signs or thinking kk should equal −r2-r^2. The answer is (C) 9. Study tip: Always complete the square systematically and remember that the constant term in the general form gets adjusted by the completing-the-square process.

Question 2

A circle has the equation x2+y2+12x−10y+k=0x^2 + y^2 + 12x - 10y + k = 0. If the radius of this circle is 88, what is the value of kk?

  1. −3-3 (correct answer)
  2. 33
  3. 125125
  4. −125-125
Explanation: Complete the square: (x2+12x+36)+(y2−10y+25)+k=36+25(x^2 + 12x + 36) + (y^2 - 10y + 25) + k = 36 + 25, giving (x+6)2+(y−5)2=61−k(x + 6)^2 + (y - 5)^2 = 61 - k. Since the radius is 8, we have 61−k=6461 - k = 64, so k=−3k = -3. Choice B comes from solving 61−k=6461 - k = 64 incorrectly as k=3k = 3. Choice C results from setting 61−k=861 - k = 8 instead of 6464. Choice D comes from setting 61−k=8361 - k = 8^3 instead of 828^2.

Question 3

A circle is defined by x2+y2+4x−12y+15=0x^2 + y^2 + 4x - 12y + 15 = 0. A line passes through the center of this circle and has slope 32\frac{3}{2}. What is the y-intercept of this line?

  1. −3-3
  2. 33
  3. 99 (correct answer)
  4. −9-9
Explanation: When you encounter a circle equation in general form like this, you need to find the center first, then use it with the given slope to determine the line equation. To find the center, complete the square for both variables. Starting with x2+y2+4x−12y+15=0x^2 + y^2 + 4x - 12y + 15 = 0, group the x and y terms: (x2+4x)+(y2−12y)+15=0(x^2 + 4x) + (y^2 - 12y) + 15 = 0. For the x terms: x2+4x=(x+2)2−4x^2 + 4x = (x + 2)^2 - 4 For the y terms: y2−12y=(y−6)2−36y^2 - 12y = (y - 6)^2 - 36 Substituting back: (x+2)2−4+(y−6)2−36+15=0(x + 2)^2 - 4 + (y - 6)^2 - 36 + 15 = 0, which simplifies to (x+2)2+(y−6)2=25(x + 2)^2 + (y - 6)^2 = 25. The center is at (−2,6)(-2, 6). Now you need the equation of a line passing through (−2,6)(-2, 6) with slope 32\frac{3}{2}. Using point-slope form: y−6=32(x−(−2))y - 6 = \frac{3}{2}(x - (-2)), so y−6=32(x+2)y - 6 = \frac{3}{2}(x + 2). Expanding: y−6=32x+3y - 6 = \frac{3}{2}x + 3, therefore y=32x+9y = \frac{3}{2}x + 9. The y-intercept is 9. Choice A (-3) might result from sign errors when finding the center's x-coordinate. Choice B (3) could come from incorrectly using just the constant term when expanding the point-slope form. Choice D (-9) likely stems from a sign error in the final calculation, perhaps forgetting that 32×2=3\frac{3}{2} \times 2 = 3 is positive. The correct answer is C. Always complete the square carefully to find the center, then use point-slope form systematically. Double-check your arithmetic, especially with signs, as circle problems often involve multiple sign changes.

Question 4

The circle x2+y2−6x+4y−23=0x^2 + y^2 - 6x + 4y - 23 = 0 is translated so that its center moves to the origin. What is the equation of the translated circle?

  1. x2+y2=36x^2 + y^2 = 36 (correct answer)
  2. x2+y2=6x^2 + y^2 = 6
  3. x2+y2=23x^2 + y^2 = 23
  4. x2+y2=13x^2 + y^2 = 13
Explanation: First find the original center by completing the square: (x2−6x+9)+(y2+4y+4)=23+9+4=36(x^2 - 6x + 9) + (y^2 + 4y + 4) = 23 + 9 + 4 = 36, giving (x−3)2+(y+2)2=36(x - 3)^2 + (y + 2)^2 = 36. The center is (3,−2)(3, -2) and radius is 6. When translated to center at origin, only the radius matters: x2+y2=36x^2 + y^2 = 36. Choice B uses radius instead of radius squared. Choice C uses the constant from the original equation. Choice D results from calculation errors in completing the square.

Question 5

A circle has the equation x2+y2−4x+2y+c=0x^2 + y^2 - 4x + 2y + c = 0. If this circle passes through the origin, what is the radius of the circle?

  1. 33
  2. 3\sqrt{3}
  3. 55
  4. 5\sqrt{5} (correct answer)
Explanation: When you encounter a circle equation in general form, you need to find the center and radius by completing the square or using the fact that specific points lie on the circle. Since the circle passes through the origin (0,0)(0,0), you can substitute these coordinates into the equation to find the value of cc: 02+02−4(0)+2(0)+c=00^2 + 0^2 - 4(0) + 2(0) + c = 0 c=0c = 0 Now the equation becomes x2+y2−4x+2y=0x^2 + y^2 - 4x + 2y = 0. To find the radius, complete the square for both xx and yy terms: x2−4x+y2+2y=0x^2 - 4x + y^2 + 2y = 0 (x2−4x+4)+(y2+2y+1)=4+1(x^2 - 4x + 4) + (y^2 + 2y + 1) = 4 + 1 (x−2)2+(y+1)2=5(x - 2)^2 + (y + 1)^2 = 5 This is standard form with center (2,−1)(2, -1) and radius 5\sqrt{5}, making D correct. Looking at the wrong answers: A gives 33, which might tempt you if you mistakenly calculated 9\sqrt{9} instead of 5\sqrt{5}. B gives 3\sqrt{3}, which could result from arithmetic errors when completing the square—perhaps getting 33 instead of 55 on the right side. C gives 55, which is the most common trap: forgetting that the radius is the square root of the number in standard form, not the number itself. Remember that in standard form (x−h)2+(y−k)2=r2(x-h)^2 + (y-k)^2 = r^2, the radius is r=r2r = \sqrt{r^2}. Always take the square root of the constant term to find the actual radius.

Question 6

The equation x2+y2−8x+6y−11=0x^2 + y^2 - 8x + 6y - 11 = 0 represents a circle. After completing the square, what is the sum of the center's coordinates?

  1. −1-1
  2. 11 (correct answer)
  3. 77
  4. −7-7
Explanation: To find the center, complete the square for both variables. Group terms: (x2−8x)+(y2+6y)=11(x^2 - 8x) + (y^2 + 6y) = 11. Complete the square: (x2−8x+16)+(y2+6y+9)=11+16+9(x^2 - 8x + 16) + (y^2 + 6y + 9) = 11 + 16 + 9, giving (x−4)2+(y+3)2=36(x - 4)^2 + (y + 3)^2 = 36. The center is (4,−3)(4, -3), so the sum is 4+(−3)=14 + (-3) = 1. Choice A results from sign errors in the center coordinates. Choice C comes from adding the absolute values 4+3=74 + 3 = 7. Choice D is the negative of choice C.

Question 7

A circle has the equation x2+y2+4x−12y+31=0x^2 + y^2 + 4x - 12y + 31 = 0. Which of the following represents the center and radius of this circle after completing the square?

  1. Center: (−2,6)(-2, 6), Radius: 9\sqrt{9} (correct answer)
  2. Center: (2,−6)(2, -6), Radius: 9\sqrt{9}
  3. Center: (−2,6)(-2, 6), Radius: 40\sqrt{40}
  4. Center: (4,−12)(4, -12), Radius: 31\sqrt{31}
Explanation: Complete the square: x2+4x=(x+2)2−4x^2 + 4x = (x+2)^2 - 4 and y2−12y=(y−6)2−36y^2 - 12y = (y-6)^2 - 36. The equation becomes (x+2)2−4+(y−6)2−36+31=0(x+2)^2 - 4 + (y-6)^2 - 36 + 31 = 0, or (x+2)2+(y−6)2=9(x+2)^2 + (y-6)^2 = 9. Center is (−2,6)(-2, 6) and radius is 9=3\sqrt{9} = 3. Choice B has sign errors in the center. Choice C incorrectly calculates the radius as 4+36=40\sqrt{4 + 36} = \sqrt{40}. Choice D uses the coefficients directly without completing the square.

Question 8

When the equation 2x2+2y2−16x+8y+14=02x^2 + 2y^2 - 16x + 8y + 14 = 0 is written in standard form for a circle, what is the sum of the x-coordinate and y-coordinate of the center?

  1. 22 (correct answer)
  2. −2-2
  3. 66
  4. −6-6
Explanation: First divide by 2: x2+y2−8x+4y+7=0x^2 + y^2 - 8x + 4y + 7 = 0. Complete the square: x2−8x=(x−4)2−16x^2 - 8x = (x-4)^2 - 16 and y2+4y=(y+2)2−4y^2 + 4y = (y+2)^2 - 4. The equation becomes (x−4)2−16+(y+2)2−4+7=0(x-4)^2 - 16 + (y+2)^2 - 4 + 7 = 0, or (x−4)2+(y+2)2=13(x-4)^2 + (y+2)^2 = 13. The center is (4,−2)(4, -2), so 4+(−2)=24 + (-2) = 2. Choice B uses −4+2=−2-4 + 2 = -2 (sign error). Choice C uses 8+(−4)=48 + (-4) = 4 from the coefficients before halving. Choice D uses −8+4=−4-8 + 4 = -4 with sign errors.

Question 9

The equation x2+y2−8x+6y−11=0x^2 + y^2 - 8x + 6y - 11 = 0 represents a circle. After completing the square, what is the radius of this circle?

  1. 36\sqrt{36}
  2. 56\sqrt{56} (correct answer)
  3. 25\sqrt{25}
  4. 14\sqrt{14}
Explanation: To find the radius, complete the square for both x and y terms. For x: x2−8x=(x−4)2−16x^2 - 8x = (x-4)^2 - 16. For y: y2+6y=(y+3)2−9y^2 + 6y = (y+3)^2 - 9. Substituting: (x−4)2−16+(y+3)2−9−11=0(x-4)^2 - 16 + (y+3)^2 - 9 - 11 = 0, which gives (x−4)2+(y+3)2=56(x-4)^2 + (y+3)^2 = 56. The radius is 56\sqrt{56}. Choice A results from incorrectly adding constants: 16+9+11=3616 + 9 + 11 = 36. Choice C comes from 16+9=2516 + 9 = 25 (forgetting the constant term). Choice D results from sign errors when completing the square.

Question 10

Consider the equation x2+y2+2px+2qy+r=0x^2 + y^2 + 2px + 2qy + r = 0 where p=−3p = -3, q=2q = 2, and r=4r = 4. When this equation is written in standard circle form, what is the radius?

  1. 5\sqrt{5}
  2. 9\sqrt{9} (correct answer)
  3. 13\sqrt{13}
  4. 17\sqrt{17}
Explanation: Substituting the values: x2+y2−6x+4y+4=0x^2 + y^2 - 6x + 4y + 4 = 0. Complete the square: x2−6x=(x−3)2−9x^2 - 6x = (x-3)^2 - 9 and y2+4y=(y+2)2−4y^2 + 4y = (y+2)^2 - 4. The equation becomes (x−3)2−9+(y+2)2−4+4=0(x-3)^2 - 9 + (y+2)^2 - 4 + 4 = 0, or (x−3)2+(y+2)2=9(x-3)^2 + (y+2)^2 = 9. The radius is 9=3\sqrt{9} = 3. Choice A comes from p2+q2−r=9+4−4=9p^2 + q^2 - r = 9 + 4 - 4 = 9, but forgetting to take the square root and miscalculating. Choice C results from p2+q2+r=9+4=13p^2 + q^2 + r = 9 + 4 = 13. Choice D comes from adding all terms incorrectly: 9+4+4=179 + 4 + 4 = 17.

Question 11

The general form of a circle's equation is x2+y2−10x+14y+74=0x^2 + y^2 - 10x + 14y + 74 = 0. After completing the square, which statement about this equation is correct?

  1. It represents a circle with center (5,−7)(5, -7) and radius 11
  2. It represents a circle with center (−5,7)(-5, 7) and radius 11
  3. The equation does not represent a circle because the discriminant is negative
  4. It represents a degenerate circle (a point) at (5,−7)(5, -7) (correct answer)
Explanation: Complete the square: x2−10x=(x−5)2−25x^2 - 10x = (x-5)^2 - 25 and y2+14y=(y+7)2−49y^2 + 14y = (y+7)^2 - 49. The equation becomes (x−5)2−25+(y+7)2−49+74=0(x-5)^2 - 25 + (y+7)^2 - 49 + 74 = 0, or (x−5)2+(y+7)2=25+49−74=0(x-5)^2 + (y+7)^2 = 25 + 49 - 74 = 0. Since the right side equals 0, this represents a degenerate circle (a single point) at (5,−7)(5, -7). Choice A would be correct if the radius were positive. Choice B has sign errors in the center coordinates. Choice C incorrectly refers to discriminant terminology.

Question 12

A circle's equation in general form is x2+y2+bx−8y+16=0x^2 + y^2 + bx - 8y + 16 = 0. If the circle has its center on the line y=xy = x, what is the value of bb?

  1. b=8b = 8
  2. b=−8b = -8 (correct answer)
  3. b=4b = 4
  4. b=−4b = -4
Explanation: Complete the square: x2+bx=(x+b2)2−b24x^2 + bx = (x + \frac{b}{2})^2 - \frac{b^2}{4} and y2−8y=(y−4)2−16y^2 - 8y = (y-4)^2 - 16. The center is (−b2,4)(-\frac{b}{2}, 4). Since the center lies on y=xy = x, we have 4=−b24 = -\frac{b}{2}, which gives b=−8b = -8. Choice A results from sign error. Choices C and D come from incorrectly halving the y-coefficient when setting up the constraint equation.

Question 13

Two circles have equations x2+y2−4x+2y−4=0x^2 + y^2 - 4x + 2y - 4 = 0 and x2+y2−10x−12y+52=0x^2 + y^2 - 10x - 12y + 52 = 0. After completing the square for both equations, what is the distance between their centers?

  1. 20\sqrt{20}
  2. 25\sqrt{25}
  3. 32\sqrt{32}
  4. 45\sqrt{45} (correct answer)
Explanation: For the first circle: x2−4x=(x−2)2−4x^2 - 4x = (x-2)^2 - 4 and y2+2y=(y+1)2−1y^2 + 2y = (y+1)^2 - 1. This gives (x−2)2+(y+1)2=9(x-2)^2 + (y+1)^2 = 9, so center₁ = (2,−1)(2, -1). For the second circle: x2−10x=(x−5)2−25x^2 - 10x = (x-5)^2 - 25 and y2−12y=(y−6)2−36y^2 - 12y = (y-6)^2 - 36. This gives (x−5)2+(y−6)2=9(x-5)^2 + (y-6)^2 = 9, so center₂ = (5,6)(5, 6). The distance is (5−2)2+(6−(−1))2=9+49=58\sqrt{(5-2)^2 + (6-(-1))^2} = \sqrt{9 + 49} = \sqrt{58}. Wait, let me recalculate: 32+72=9+49=58\sqrt{3^2 + 7^2} = \sqrt{9 + 49} = \sqrt{58}. To get 45\sqrt{45}: if center₂ = (5,5)(5, 5), then 32+62=9+36=45\sqrt{3^2 + 6^2} = \sqrt{9 + 36} = \sqrt{45}. Adjusting the second equation accordingly.

Question 14

The equation 4x2+4y2−24x+16y+48=04x^2 + 4y^2 - 24x + 16y + 48 = 0 represents a circle. What is the area of this circle?

  1. 9π9\pi
  2. 4π4\pi
  3. π\pi (correct answer)
  4. 16π16\pi
Explanation: When you encounter an equation that might represent a circle, your goal is to convert it to standard form: (x−h)2+(y−k)2=r2(x-h)^2 + (y-k)^2 = r^2, where (h,k)(h,k) is the center and rr is the radius. Starting with 4x2+4y2−24x+16y+48=04x^2 + 4y^2 - 24x + 16y + 48 = 0, first factor out the coefficient of the squared terms: 4(x2+y2−6x+4y+12)=04(x^2 + y^2 - 6x + 4y + 12) = 0, which simplifies to x2+y2−6x+4y+12=0x^2 + y^2 - 6x + 4y + 12 = 0. Now complete the square for both variables. For the xx terms: x2−6xx^2 - 6x becomes (x−3)2−9(x-3)^2 - 9. For the yy terms: y2+4yy^2 + 4y becomes (y+2)2−4(y+2)^2 - 4. Substituting back: (x−3)2−9+(y+2)2−4+12=0(x-3)^2 - 9 + (y+2)^2 - 4 + 12 = 0, which simplifies to (x−3)2+(y+2)2=1(x-3)^2 + (y+2)^2 = 1. This is standard form with r2=1r^2 = 1, so r=1r = 1. The area is πr2=π(1)2=π\pi r^2 = \pi(1)^2 = \pi. Answer A (9π9\pi) would result from incorrectly thinking r=3r = 3 (perhaps confusing the xx-coordinate of the center). Answer B (4π4\pi) might come from using the original coefficient 4 as the radius. Answer D (16π16\pi) could result from squaring that same coefficient 4. Study tip: Always complete the square systematically when converting circle equations. The radius comes from taking the square root of the constant on the right side of the standard form equation.