Algebra 3 Quiz: Circles And Parabolas
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Circles And ParabolasQuestion 1 of 12

A circle has a diameter whose endpoints are (1,−2)(1,-2) and (7,4)(7,4). Which equation represents this circle?

(x−4)2+(y−1)2=18(x-4)^2+(y-1)^2=18
(x−4)2+(y−1)2=72(x-4)^2+(y-1)^2=72
(x−4)2+(y−1)2=36(x-4)^2+(y-1)^2=36
(x+4)2+(y+1)2=18(x+4)^2+(y+1)^2=18
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Algebra 3 Quiz

Algebra 3 Quiz: Circles And Parabolas

Practice Circles And Parabolas in Algebra 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 Circles And Parabolas, giving you a quick way to practice the rules, question types, and explanations that matter most for Algebra 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

A circle has a diameter whose endpoints are (1,−2)(1,-2) and (7,4)(7,4). Which equation represents this circle?

  1. (x−4)2+(y−1)2=18(x-4)^2+(y-1)^2=18 (correct answer)
  2. (x−4)2+(y−1)2=72(x-4)^2+(y-1)^2=72
  3. (x−4)2+(y−1)2=36(x-4)^2+(y-1)^2=36
  4. (x+4)2+(y+1)2=18(x+4)^2+(y+1)^2=18
Explanation: Whenever you see a circle defined by a diameter, your first move is to find the center and the radius. The center is the midpoint of the endpoints: (1+72,−2+42)=(4,1)\left(\frac{1+7}{2},\frac{-2+4}{2}\right)=(4,1). To get the radius, find the distance from the center to one endpoint: (7−4)2+(4−1)2=9+9=18\sqrt{(7-4)^2+(4-1)^2}=\sqrt{9+9}=\sqrt{18}. Thus the equation is (x−4)2+(y−1)2=18(x-4)^2+(y-1)^2=18, since the standard form uses (x−h)2+(y−k)2=r2(x-h)^2+(y-k)^2=r^2. The wrong choices show common traps. (x−4)2+(y−1)2=72(x-4)^2+(y-1)^2=72 uses the squared length of the entire diameter — the distance between the endpoints is 72\sqrt{72}, and that is 4r24r^2, not r2r^2. (x−4)2+(y−1)2=36(x-4)^2+(y-1)^2=36 comes from dividing that squared diameter by 2 instead of 4, giving a radius of 6 rather than 18\sqrt{18}. Finally, (x+4)2+(y+1)2=18(x+4)^2+(y+1)^2=18 has the correct radius but the center (−4,−1)(-4,-1); that sign error usually comes from miscomputing the midpoint. For the exam, follow a routine: midpoint for the center, then distance from the center to either endpoint for the radius. Always square that radius carefully — and never confuse it with the diameter.

Question 2

A parabola has vertex (3,−2)(3,-2) and focus (3,1)(3,1). Which equation represents the parabola?

  1. (x−3)2=−12(y+2)(x-3)^2=-12(y+2)
  2. (y+2)2=12(x−3)(y+2)^2=12(x-3)
  3. (x−3)2=12(y+2)(x-3)^2=12(y+2) (correct answer)
  4. (x−3)2=6(y+2)(x-3)^2=6(y+2)
Explanation: When you see a parabola with a vertex and focus, first determine which way it opens by comparing the coordinates. Here both have x=3x=3, so the axis is vertical; since the focus (3,1)(3,1) is above the vertex (3,−2)(3,-2), the parabola opens upward. For a vertical parabola, the standard form is (x−h)2=4p(y−k)(x-h)^2=4p(y-k), where (h,k)(h,k) is the vertex and pp is the directed distance from vertex to focus. The distance from −2-2 to 11 is 33, so p=3p=3, making 4p=124p=12. Substituting h=3h=3, k=−2k=-2, you get: (x−3)2=12(y+2)(x-3)^2=12(y+2) That is the correct equation. Now look at the traps. (x−3)2=−12(y+2)(x-3)^2=-12(y+2) would have p=−3p=-3, meaning the parabola opens downward — not what the focus indicates. (y+2)2=12(x−3)(y+2)^2=12(x-3) is a horizontal parabola opening right; its focus would be at (6,−2)(6,-2), not (3,1)(3,1). Finally, (x−3)2=6(y+2)(x-3)^2=6(y+2) uses 4p=64p=6, so p=1.5p=1.5, which is only half the actual vertex-to-focus distance. A good habit: always compute pp as the distance from vertex to focus, then multiply by 44. Also match the squared variable to the axis — if the focus changes yy, the parabola opens vertically, so xx is squared.

Question 3

A parabola is given by y2−6y+8x+1=0y^2-6y+8x+1=0. What are its focus and directrix?

  1. Focus (3,3)(3,3); directrix x=−1x=-1
  2. Focus (−3,3)(-3,3); directrix x=5x=5
  3. Focus (1,3)(1,3); directrix x=3x=3
  4. Focus (−1,3)(-1,3); directrix x=3x=3 (correct answer)
Explanation: When you see a parabola with the yy term squared, it opens left or right. Your first move is to complete the square to rewrite it in the standard form (y−k)2=4p(x−h)(y-k)^2 = 4p(x-h). Here, y2−6y+8x+1=0y^2-6y+8x+1=0 becomes y2−6y=−8x−1y^2-6y = -8x-1. Completing the square on the left: (y−3)2−9=−8x−1(y-3)^2 - 9 = -8x -1, so (y−3)2=−8x+8=−8(x−1)(y-3)^2 = -8x +8 = -8(x-1). Thus h=1h=1, k=3k=3, and 4p=−84p=-8 so p=−2p=-2. The focus is (h+p,k)=(1−2,3)=(−1,3)(h+p,k) = (1-2,3) = (-1,3), and the directrix is x=h−p=1−(−2)=3x = h-p = 1-(-2) = 3. That matches the choice "Focus (−1,3)(-1,3); directrix x=3x=3". Now, why the others are traps. "Focus (3,3)(3,3); directrix x=−1x=-1" uses p=+2p=+2, which would come from (y−3)2=8(x−1)(y-3)^2 = 8(x-1) — the sign error of ignoring the negative coefficient. "Focus (−3,3)(-3,3); directrix x=5x=5" gives a vertex at x=1x=1 but a distance p=−4p=-4 (or +4+4), double the correct magnitude, likely from misreading 4p=−84p=-8 as p=−4p=-4. Finally, "Focus (1,3)(1,3); directrix x=3x=3" confuses the vertex (1,3)(1,3) with the focus; the vertex is halfway between focus and directrix, so the focus would be (−1,3)(-1,3) if the directrix is x=3x=3. Your key check: after completing the square, compare 4p4p carefully — the sign tells you which way the parabola opens, and the distance ∣p∣|p| sets the focus and directrix. Always compute h+ph+p and h−ph-p directly.

Question 4

The circle x2+y2−4x+8y+11=0x^2+y^2-4x+8y+11=0 is graphed. Which point lies outside the circle?

  1. (2,−1)(2,-1)
  2. (4,−4)(4,-4)
  3. (5,−3)(5,-3) (correct answer)
  4. (0,−2)(0,-2)
Explanation: Whenever you see a circle equation like this, your first move is to rewrite it in standard form by completing the square — that reveals the center and radius, which are the keys to deciding whether a point is inside, on, or outside the circle. Complete the square:
x2−4x+y2+8y=−11x^2-4x + y^2+8y = -11
(x−2)2−4+(y+4)2−16=−11(x-2)^2 -4 + (y+4)^2 -16 = -11
(x−2)2+(y+4)2=9(x-2)^2 + (y+4)^2 = 9
So the center is (2,−4)(2,-4) and the radius is 33. A point lies outside if its distance from the center is greater than 33. Check each candidate.
  • (2,−1)(2,-1): distance from (2,−4)(2,-4) is ∣3∣=3|3| = 3 — it lies on the circle, not outside.
  • (4,−4)(4,-4): distance is (2)2+02=2\sqrt{(2)^2+0^2}=2 — inside.
  • (5,−3)(5,-3): distance is (3)2+(1)2=10≈3.16\sqrt{(3)^2+(1)^2}=\sqrt{10}\approx 3.16 — greater than 33, so this point is outside.
  • (0,−2)(0,-2): distance is (−2)2+(2)2=8≈2.83\sqrt{(-2)^2+(2)^2}=\sqrt{8}\approx 2.83 — inside.
The trap here is confusing "on" with "outside" — a point exactly at distance 33 is not outside. Also, make sure you complete the square correctly; the center is easy to misread if you forget to adjust the constant. Study tip: For any circle, always find the center and radius first. Then test points by computing their distance from the center — no need to plug into the original equation. Memorize the distance comparison: <r<r inside, =r=r on, >r>r outside.

Question 5

A circle passes through (−2,0)(-2,0) and (6,0)(6,0),and its center lies on the line y=xy=x. Which equation represents this circle?

  1. (x−2)2+(y−2)2=4(x-2)^2+(y-2)^2=4
  2. (x−2)2+(y+2)2=20(x-2)^2+(y+2)^2=20
  3. (x−2)2+(y−2)2=16(x-2)^2+(y-2)^2=16
  4. (x−2)2+(y−2)2=20(x-2)^2+(y-2)^2=20 (correct answer)
Explanation: When you see a circle question with two given points, think geometrically: the center must be equidistant from both points, so it lies on the perpendicular bisector of the segment joining them. Here the points are (−2,0)(-2,0) and (6,0)(6,0), a horizontal segment with midpoint (2,0)(2,0), so the perpendicular bisector is x=2x=2. Since the center also lies on y=xy=x, the center must be (2,2)(2,2). The radius is the distance from (2,2)(2,2) to either point, say (6,0)(6,0): (6−2)2+(0−2)2=20\sqrt{(6-2)^2+(0-2)^2}=\sqrt{20}. Thus the circle's equation is (x−2)2+(y−2)2=20(x-2)^2+(y-2)^2=20. The choice (x−2)2+(y−2)2=4(x-2)^2+(y-2)^2=4 has the correct center but radius 22, which is too small to reach either given point. The choice (x−2)2+(y+2)2=20(x-2)^2+(y+2)^2=20 has center (2,−2)(2,-2), which is not on y=xy=x; the plus sign in y+2y+2 means the center's yy-coordinate is −2-2. The choice (x−2)2+(y−2)2=16(x-2)^2+(y-2)^2=16 uses radius 44, a common trap: it treats the segment between (−2,0)(-2,0) and (6,0)(6,0) as a diameter, but those are just two points on the circle, not opposite endpoints of a diameter. A strong strategy: when a circle passes through two points, immediately find the perpendicular bisector of those points. Its intersection with any other condition gives the center. Then calculate the radius by distance to one of the points. Never assume the given points are endpoints of a diameter unless the center is exactly their midpoint.

Question 6

A parabola is given by x=12(y−1)2+3x=\frac{1}{2}(y-1)^2+3. What are its vertex and focus?

  1. Vertex (3,1)(3,1); focus (3,32)(3,\frac{3}{2})
  2. Vertex (3,1)(3,1); focus (52,1)(\frac{5}{2},1)
  3. Vertex (3,1)(3,1); focus (72,1)(\frac{7}{2},1) (correct answer)
  4. Vertex (−3,−1)(-3,-1); focus (72,−1)(\frac{7}{2},-1)
Explanation: Whenever you see a parabola written as x=a(y−k)2+hx=a(y-k)^2+h, the roles of xx and yy are swapped compared with y=a(x−h)2+ky=a(x-h)^2+k. The vertex is (h,k)(h,k), and the parabola opens right if a>0a>0, left if a<0a<0. Here h=3h=3 and k=1k=1, so the vertex is (3,1)(3,1). Rewrite the equation to find the focal length: x=12(y−1)2+3  ⟹  (y−1)2=2(x−3).x=\frac12(y-1)^2+3 \implies (y-1)^2=2(x-3). This matches (y−k)2=4p(x−h)(y-k)^2=4p(x-h), so 4p=24p=2, giving p=12p=\frac12. Since the parabola opens right, the focus is 12\frac12 unit to the right of the vertex: (3+12,1)=(72,1).(3+\tfrac12,1)=\left(\frac72,1\right). So the vertex (3,1)(3,1); focus (72,1)(\frac72,1) is the correct pairing. The focus (3,32)(3,\frac32) treats the parabola as if it opened upward, but since yy is the squared variable, the focus must move horizontally, not vertically. The focus (52,1)(\frac52,1) is actually the directrix line, which lies 12\frac12 unit to the left of the vertex. The vertex (−3,−1)(-3,-1) uses the opposite signs for both the horizontal and vertical shifts; +3+3 and −1-1 in the equation mean (3,1)(3,1), not (−3,−1)(-3,-1). Its focus also has the wrong xx-coordinate. On the exam, when a parabola has xx isolated and yy squared, think "sideways": vertex and focus share the same yy-coordinate, and the focus is pp units right or left of the vertex.

Question 7

A circle is centered at (1,−3)(1,-3) and is tangent to the line 4x−3y=74x-3y=7. Which equation represents this circle?

  1. (x−1)2+(y+3)2=36(x-1)^2+(y+3)^2=36
  2. (x−1)2+(y+3)2=3625(x-1)^2+(y+3)^2=\frac{36}{25} (correct answer)
  3. (x−1)2+(y+3)2=65(x-1)^2+(y+3)^2=\frac{6}{5}
  4. (x+1)2+(y−3)2=3625(x+1)^2+(y-3)^2=\frac{36}{25}
Explanation: When you see a circle tangent to a line, the key fact is that the radius equals the perpendicular distance from the center to that tangent line. So here, start with the center (1,−3)(1,-3), which gives the circle equation form (x−1)2+(y+3)2=r2(x-1)^2+(y+3)^2=r^2. To find rr, use the distance from a point to a line: ∣4(1)−3(−3)−7∣42+(−3)2=∣4+9−7∣5=65\frac{|4(1)-3(-3)-7|}{\sqrt{4^2+(-3)^2}}=\frac{|4+9-7|}{5}=\frac{6}{5}. That distance is the radius, so the right side must be r2r^2, not rr: (65)2=3625\left(\frac{6}{5}\right)^2=\frac{36}{25}. Thus the correct equation is (x−1)2+(y+3)2=3625(x-1)^2+(y+3)^2=\frac{36}{25}. Now look at the traps in the other choices. The equation with (x−1)2+(y+3)2=36(x-1)^2+(y+3)^2=36 uses the correct center but treats 6/56/5 as if it were 66, forgetting to divide by the square-root denominator in the distance formula. The equation with (x-1)^2+(y+3)^2=\frac{6}{5}} uses the correct center but puts the radius itself on the right instead of the radius squared; the right side of a circle equation is always r2r^2. The equation with (x+1)2+(y−3)2=3625(x+1)^2+(y-3)^2=\frac{36}{25} has the correct radius squared but the wrong center signs — it corresponds to center (−1,3)(-1,3), not (1,−3)(1,-3). Your study takeaway: whenever a circle is tangent to a line, immediately set radius = distance from center to line, then square that distance for the equation. Also double-check that the signs inside (x−h)(x-h) and (y−k)(y-k) match the center coordinates.

Question 8

The vertex of a downward-opening parabola is (−1,4)(-1,4), and one xx-intercept is 33. What is the other xx-intercept?

  1. −3-3
  2. 11
  3. 55
  4. −5-5 (correct answer)
Explanation: When you see a parabola with a given vertex and one xx-intercept, the key is to remember that the axis of symmetry passes through the vertex. For any parabola, the two xx-intercepts are mirror images across that vertical line. Here the vertex is (−1,4)(-1,4), so the axis of symmetry is x=−1x=-1. Your known intercept is 33. The distance from the axis to that intercept is 3−(−1)=43 - (-1) = 4. To find the mirror point, move the same distance to the other side: −1−4=−5-1 - 4 = -5. That is the other xx-intercept, so the correct choice is −5-5. Now look at the wrong answers. −3-3 would be the intercept if you mistakenly used the vertex's xx-coordinate as the midpoint between 33 and itself, but the midpoint of 33 and −3-3 is 00, not −1-1. 11 is too close—the distance from −1-1 to 11 is only 22, so it would not balance the 44-unit distance on the right. 55 is on the correct side but overshoots; its distance from −1-1 is 66, not 44. These all fail to respect the symmetry about the vertex's xx-coordinate. A quick strategy: whenever a parabola's vertex is given, write down the axis of symmetry x=hx = h. Then for any known xx-intercept rr, the other intercept is 2h−r2h - r. That formula saves time and prevents mistakes. On the exam, always verify the axis before assuming symmetry about 00.

Question 9

For which value(s) of kk does x2+y2+6x−4y+k=0x^2+y^2+6x-4y+k=0 represent a circle?

  1. k>13k>13
  2. k<13k<13 (correct answer)
  3. k≤13k\le 13
  4. k=13k=13
Explanation: Whenever you see a general second-degree equation like x2+y2+6x−4y+k=0x^2+y^2+6x-4y+k=0, your first move should be to complete the square to reveal the circle's center and radius. Group the xx terms and yy terms: (x2+6x)+(y2−4y)+k=0(x^2+6x)+(y^2-4y)+k=0. Completing the square gives (x+3)2−9+(y−2)2−4+k=0(x+3)^2-9+(y-2)^2-4+k=0, so (x+3)2+(y−2)2=13−k(x+3)^2+(y-2)^2=13-k. This is a circle exactly when the right side is positive, because the radius squared must be greater than zero. Thus you need 13−k>013-k>0, which simplifies to k<13k<13. Now look at the other choices. The choice k>13k>13 would make 13−k13-k negative, so no real points satisfy the equation — that gives an empty graph, not a circle. The choice k≤13k\le 13 includes too much: when k=13k=13, the equation becomes (x+3)2+(y−2)2=0(x+3)^2+(y-2)^2=0, which is only the single point (−3,2)(-3,2), not a circle. The choice k=13k=13 is exactly that degenerate point case, so it fails the requirement of having a positive radius. Your takeaway: after completing the square, always check the radius-squared constant. For a circle, it must be strictly positive. If it equals zero, you get a point; if it is negative, there is no graph. On this exam, the trap is forgetting that k=13k=13 produces a degenerate case, so remember the strict inequality.

Question 10

A parabola has focus (2,−3)(2,-3) and directrix y=5y=5. Which equation represents the parabola?

  1. (x−2)2=−16(y−1)(x-2)^2=-16(y-1) (correct answer)
  2. (x−2)2=16(y−1)(x-2)^2=16(y-1)
  3. (y−1)2=−16(x−2)(y-1)^2=-16(x-2)
  4. (x−2)2=−8(y−1)(x-2)^2=-8(y-1)
Explanation: When you see a focus and directrix, remember that the vertex is the midpoint between them, and the parabola opens away from the directrix. Here the focus is (2,−3)(2,-3) and the directrix is y=5y=5, so the axis is vertical. The midpoint on x=2x=2 has y=−3+52=1y=\frac{-3+5}{2}=1, giving vertex (2,1)(2,1). Since the focus lies below the vertex, p=−4p=-4, and the standard form is (x−h)2=4p(y−k)(x-h)^2=4p(y-k). Substituting gives (x−2)2=−16(y−1)(x-2)^2=-16(y-1), so that is the correct equation. The form (x−2)2=16(y−1)(x-2)^2=16(y-1) has the same vertex but uses p=+4p=+4, which would make the parabola open upward toward the directrix — opposite of the given focus/directrix setup. The form (y−1)2=−16(x−2)(y-1)^2=-16(x-2) represents a sideways-opening parabola, which would require a vertical directrix, not the horizontal line y=5y=5. The form (x−2)2=−8(y−1)(x-2)^2=-8(y-1) gives only 4p=−84p=-8, so p=−2p=-2, but the true distance from the vertex to the focus is 44, not 22. Quick strategy: always locate the vertex and decide the axis direction first. If the directrix is horizontal, the equation is of the form (x−h)2=4p(y−k)(x-h)^2=4p(y-k); if the directrix is vertical, it is (y−k)2=4p(x−h)(y-k)^2=4p(x-h). The sign of pp tells you which way the parabola opens.

Question 11

What is the circumference of the circle given by x2+y2+6x−10y−15=0x^2+y^2+6x-10y-15=0?

  1. 7π7\pi
  2. 14π14\pi (correct answer)
  3. 49π49\pi
  4. 98π98\pi
Explanation: Whenever you see a circle equation in expanded form, your first move is to complete the square to uncover the center and radius. For x2+y2+6x−10y−15=0x^2+y^2+6x-10y-15=0, group the xx terms and the yy terms: (x2+6x)+(y2−10y)=15(x^2+6x)+(y^2-10y)=15. Complete each square: (x+3)2−9+(y−5)2−25=15(x+3)^2-9+(y-5)^2-25=15, so (x+3)2+(y−5)2=49(x+3)^2+(y-5)^2=49. The radius is 49=7\sqrt{49}=7, and circumference is 2πr=14π2\pi r = 14\pi. That is the correct choice. The distractor 7π7\pi is the radius itself, not the circumference — it forgets the factor of 2π2\pi. The choice 49π49\pi is the area of the circle, πr2\pi r^2, not its perimeter. The choice 98π98\pi is double the area, a sign of confusing area with circumference and then multiplying by 2. Each wrong answer comes from mixing up the circle formulas: radius, diameter, area, and circumference. On this exam, remember the key formulas: circumference is 2πr2\pi r or πd\pi d, and area is πr2\pi r^2. When the equation is given in expanded form, completing the square is always the first step — then read the radius straight from the standard form. A quick check: if the constant on the right is a perfect square like 4949, take its square root before doing anything with π\pi.

Question 12

A vertical-axis parabola has vertex (2,5)(2,5) and passes through (4,3)(4,3). What is its focus?

  1. (2,92)(2,\frac{9}{2}) (correct answer)
  2. (2,112)(2,\frac{11}{2})
  3. (4,3)(4,3)
  4. (2,5)(2,5)
Explanation: When you see a vertical-axis parabola with a vertex and another point, your first step is to find the focal length pp. The standard form is (x−h)2=4p(y−k)(x-h)^2 = 4p(y-k), where the vertex is (h,k)(h,k) and the focus is (h,k+p)(h, k+p). Here h=2h=2, k=5k=5. Plug in the given point (4,3)(4,3): (4−2)2=4p(3−5)(4-2)^2 = 4p(3-5), so 4=−8p4 = -8p, giving p=−12p = -\frac12. The focus is therefore (2,5−12)=(2,92)(2, 5 - \frac12) = (2, \frac{9}{2}). The negative pp tells you the parabola opens downward, which matches the point lying below the vertex. The option (2,112)(2, \frac{11}{2}) would come from mistakenly using p=+12p = +\frac12, which would open upward and place the point above the vertex—contradicting the given point. The point (4,3)(4,3) is just a point on the curve, not the focus. The vertex (2,5)(2,5) is the midpoint between focus and directrix, not the focus itself. Always check the sign of pp by comparing the given point's yy-coordinate to the vertex's yy-coordinate. For any vertical-axis parabola, remember that focus and vertex share the same xx-coordinate. Write the standard form, substitute the given point to solve for pp, and verify the direction from the sign of pp. That will lead you straight to the focus.