Math 3 Quiz: Circle Equations
4 questions · exam conditions
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Circle EquationsQuestion 1 of 4

Two concentric circles have equations x2+y2−6x+8y=0x^2 + y^2 - 6x + 8y = 0 and x2+y2−6x+8y+k=0x^2 + y^2 - 6x + 8y + k = 0. If the area of the annulus (ring) between them is 21π21\pi, what is the value of kk?

−21-21
−15-15
1515
2121
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Math 3 Quiz

Math 3 Quiz: Circle Equations

Practice Circle Equations in Math 3 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 Circle Equations, giving you a quick way to practice the rules, question types, and explanations that matter most for Math 3.

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

Two concentric circles have equations x2+y2−6x+8y=0x^2 + y^2 - 6x + 8y = 0 and x2+y2−6x+8y+k=0x^2 + y^2 - 6x + 8y + k = 0. If the area of the annulus (ring) between them is 21π21\pi, what is the value of kk?

  1. −21-21 (correct answer)
  2. −15-15
  3. 1515
  4. 2121
Explanation: Complete the square for the first circle: (x−3)2+(y+4)2=9+16=25(x-3)^2 + (y+4)^2 = 9 + 16 = 25, so r1=5r_1 = 5. For the second circle: (x−3)2+(y+4)2=25−k(x-3)^2 + (y+4)^2 = 25 - k, so r2=25−kr_2 = \sqrt{25-k}. The annulus area is πr12−πr22=π(25)−π(25−k)=πk\pi r_1^2 - \pi r_2^2 = \pi(25) - \pi(25-k) = \pi k. Setting this equal to 21π21\pi gives k=21k = 21. But this assumes r1>r2r_1 > r_2, which requires k>0k > 0. If k<0k < 0, then r2>r1r_2 > r_1 and the area is π(25−k)−25π=−πk=21π\pi(25-k) - 25\pi = -\pi k = 21\pi, so k=−21k = -21. Choice A is correct.

Question 2

A circle intersects the coordinate axes at exactly three points. If two of these points are (−4,0)(-4, 0) and (0,3)(0, 3), and the circle passes through the origin, what is the xx-coordinate of the center?

  1. −52-\frac{5}{2}
  2. −2-2 (correct answer)
  3. −32-\frac{3}{2}
  4. −1-1
Explanation: Since the circle passes through (0,0)(0,0), (−4,0)(-4,0), and (0,3)(0,3), we can use the general form x2+y2+Dx+Ey+F=0x^2 + y^2 + Dx + Ey + F = 0. From (0,0)(0,0): F=0F = 0. From (−4,0)(-4,0): 16−4D=016 - 4D = 0, so D=4D = 4. From (0,3)(0,3): 9+3E=09 + 3E = 0, so E=−3E = -3. The equation is x2+y2+4x−3y=0x^2 + y^2 + 4x - 3y = 0. Completing the square: (x+2)2+(y−32)2=4+94=254(x+2)^2 + (y-\frac{3}{2})^2 = 4 + \frac{9}{4} = \frac{25}{4}. The center is (−2,32)(-2, \frac{3}{2}), so the xx-coordinate is −2-2. Choice A uses wrong arithmetic. Choice C confuses xx and yy coordinates. Choice D is off by 1.

Question 3

A circle passes through points A(0,0)A(0, 0), B(6,0)B(6, 0), and C(3,4)C(3, 4). What is the radius of this circle?

  1. 154\frac{15}{4}
  2. 258\frac{25}{8} (correct answer)
  3. 52\frac{5}{2}
  4. 158\frac{15}{8}
Explanation: Let the circle have equation (x−h)2+(y−k)2=r2(x-h)^2 + (y-k)^2 = r^2. Using the three points: From A(0,0): h2+k2=r2h^2 + k^2 = r^2. From B(6,0): (6−h)2+k2=r2(6-h)^2 + k^2 = r^2. From C(3,4): (3−h)2+(4−k)2=r2(3-h)^2 + (4-k)^2 = r^2. Setting the first two equal: h2+k2=(6−h)2+k2h^2 + k^2 = (6-h)^2 + k^2, so h2=36−12h+h2h^2 = 36 - 12h + h^2, giving 12h=3612h = 36 and h=3h = 3. Setting the first and third equal: h2+k2=(3−h)2+(4−k)2h^2 + k^2 = (3-h)^2 + (4-k)^2. Substituting h=3h = 3: 9+k2=0+(4−k)2=16−8k+k29 + k^2 = 0 + (4-k)^2 = 16 - 8k + k^2, so 9=16−8k9 = 16 - 8k, giving 8k=78k = 7 and k=78k = \frac{7}{8}. The radius is r=h2+k2=9+4964=576+4964=62564=258r = \sqrt{h^2 + k^2} = \sqrt{9 + \frac{49}{64}} = \sqrt{\frac{576 + 49}{64}} = \sqrt{\frac{625}{64}} = \frac{25}{8}.

Question 4

A circle has equation x2+y2−8x+2y+8=0x^2 + y^2 - 8x + 2y + 8 = 0. If this circle is reflected across the line y=xy = x, what is the equation of the resulting circle?

  1. x2+y2+2x−8y+8=0x^2 + y^2 + 2x - 8y + 8 = 0 (correct answer)
  2. x2+y2−2x+8y+8=0x^2 + y^2 - 2x + 8y + 8 = 0
  3. x2+y2+8x−2y+8=0x^2 + y^2 + 8x - 2y + 8 = 0
  4. x2+y2−8x−2y+8=0x^2 + y^2 - 8x - 2y + 8 = 0
Explanation: First find the center of the original circle by completing the square: (x−4)2+(y+1)2=16+1−8=9(x-4)^2 + (y+1)^2 = 16 + 1 - 8 = 9. The center is (4,−1)(4, -1) with radius 33. When reflected across y=xy = x, the point (a,b)(a, b) becomes (b,a)(b, a). So the new center is (−1,4)(-1, 4). The radius remains 33. The new equation is (x+1)2+(y−4)2=9(x+1)^2 + (y-4)^2 = 9, which expands to x2+2x+1+y2−8y+16=9x^2 + 2x + 1 + y^2 - 8y + 16 = 9, giving x2+y2+2x−8y+8=0x^2 + y^2 + 2x - 8y + 8 = 0. Choice B has wrong signs. Choice C doesn't reflect properly. Choice D changes the wrong coefficients.