Algebra 2 Quiz: Derive The Equation Of A Circle
20 questions · exam conditions
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Derive The Equation Of A CircleQuestion 1 of 20

The general equation x2+y2−6x+8y+c=0x^2 + y^2 - 6x + 8y + c = 0 represents a circle with radius 55. What is the value of cc?

2525
−25-25
00
−50-50
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Algebra 2 Quiz

Algebra 2 Quiz: Derive The Equation Of A Circle

Practice Derive The Equation Of A Circle in Algebra 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 Derive The Equation Of A Circle, giving you a quick way to practice the rules, question types, and explanations that matter most for Algebra 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 general equation x2+y2−6x+8y+c=0x^2 + y^2 - 6x + 8y + c = 0 represents a circle with radius 55. What is the value of cc?

  1. 2525
  2. −25-25
  3. 00 (correct answer)
  4. −50-50
Explanation: When you see a circle equation in general form like this, you need to convert it to standard form (x−h)2+(y−k)2=r2(x-h)^2 + (y-k)^2 = r^2 by completing the square. This will reveal the center and radius, allowing you to find the missing parameter. Starting with x2+y2−6x+8y+c=0x^2 + y^2 - 6x + 8y + c = 0, rearrange and complete the square for both variables. For the xx terms: x2−6xx^2 - 6x becomes (x−3)2−9(x-3)^2 - 9 since (−6/2)2=9(-6/2)^2 = 9. For the yy terms: y2+8yy^2 + 8y becomes (y+4)2−16(y+4)^2 - 16 since (8/2)2=16(8/2)^2 = 16. Substituting back: (x−3)2−9+(y+4)2−16+c=0(x-3)^2 - 9 + (y+4)^2 - 16 + c = 0, which simplifies to (x−3)2+(y+4)2=25−c(x-3)^2 + (y+4)^2 = 25 - c. Since the radius is 5, we know r2=25r^2 = 25, so 25−c=2525 - c = 25. Therefore, c=0c = 0. Choice A (2525) would give you 25−25=025 - 25 = 0 on the right side, meaning the radius would be 0 - just a point, not a circle. Choice B (−25-25) would make the right side 25−(−25)=5025 - (-25) = 50, giving radius 50=52≈7.07\sqrt{50} = 5\sqrt{2} \approx 7.07, not 5. Choice D (−50-50) would create 25−(−50)=7525 - (-50) = 75, yielding radius 75=53≈8.66\sqrt{75} = 5\sqrt{3} \approx 8.66. Study tip: Always remember that completing the square introduces constants that must be balanced. The constant term in the original equation directly affects the radius through the relationship r2=(constants from completing the square)−cr^2 = (\text{constants from completing the square}) - c.

Question 2

All points (x,y)(x,y) that are 5 units from the point (1,2)(1,2) form a circle. Write the equation of this circle in standard form. (Use the Pythagorean Theorem/distance formula.)

  1. (x−2)2+(y−1)2=25(x-2)^2 + (y-1)^2 = 25
  2. (x−1)2+(y−2)2=25(x-1)^2 + (y-2)^2 = 25 (correct answer)
  3. (x+1)2+(y+2)2=25(x+1)^2 + (y+2)^2 = 25
  4. (x−1)2+(y−2)2=5(x-1)^2 + (y-2)^2 = 5
Explanation: This question tests your understanding of how to derive circle equations using the Pythagorean Theorem (distance formula) or find a circle's center and radius by completing the square. A circle is defined as all points at a fixed distance (radius r) from a center point (h, k): using the distance formula, the distance from any point (x, y) on the circle to the center is √[(x - h)² + (y - k)²] = r. Squaring both sides eliminates the radical and gives the standard form (x - h)² + (y - k)² = r². This equation comes directly from the Pythagorean Theorem applied to the right triangle formed by the horizontal distance (x - h), vertical distance (y - k), and radius r as hypotenuse! Here, points 5 units from (1, 2) satisfy (x - 1)² + (y - 2)² = 25. Choice C correctly derives the equation as (x-1)² + (y-2)² = 25. A distractor like choice A forgets to square the radius—it's r², not r! Deriving from center and radius: (1) Write the distance from general point (x, y) to center (h, k) using distance formula: √[(x - h)² + (y - k)²], (2) Set equal to radius r, (3) Square both sides to get (x - h)² + (y - k)² = r². That's it! For example, center (2, -5) and radius 6: (x - 2)² + (y - (-5))² = 6², which simplifies to (x - 2)² + (y + 5)² = 36. The Pythagorean Theorem makes circle equations! Fantastic— you're nailing the distance concept!

Question 3

A cell tower is located at (3,4)(3,4) on a coordinate grid. Its signal reaches all points within 1010 units of the tower. Write the equation of the circle representing points exactly 1010 units from the tower.

  1. (x+3)2+(y+4)2=100(x+3)^2+(y+4)^2=100
  2. (x−3)2+(y−4)2=100(x-3)^2+(y-4)^2=100 (correct answer)
  3. (x−3)2+(y+4)2=100(x-3)^2+(y+4)^2=100
  4. (x−3)2+(y−4)2=10(x-3)^2+(y-4)^2=10
Explanation: This question tests your understanding of how to derive circle equations using the Pythagorean Theorem (distance formula) or find a circle's center and radius by completing the square. A circle is defined as all points at a fixed distance (radius r) from a center point (h, k): using the distance formula, the distance from any point (x, y) on the circle to the center is √[(x - h)² + (y - k)²] = r. Squaring both sides eliminates the radical and gives the standard form (x - h)² + (y - k)² = r². This equation comes directly from the Pythagorean Theorem applied to the right triangle formed by the horizontal distance (x - h), vertical distance (y - k), and radius r as hypotenuse! For the tower at (3, 4) with reach 10 units, the equation is (x - 3)² + (y - 4)² = 10² = 100. Choice B correctly derives the equation with center (3, 4) and r² = 100. An error like in choice C might flip the signs to (x + 3) and (y + 4), but recall it's (x - h) and (y - k) for positive h and k. Deriving from center and radius: (1) Write the distance from general point (x, y) to center (h, k) using distance formula: √[(x - h)² + (y - k)²], (2) Set equal to radius r, (3) Square both sides to get (x - h)² + (y - k)² = r². That's it! For example, center (2, -5) and radius 6: (x - 2)² + (y - (-5))² = 6², which simplifies to (x - 2)² + (y + 5)² = 36. The Pythagorean Theorem makes circle equations!

Question 4

Two circles intersect at points AA and BB. The first circle has center (1,2)(1, 2) and passes through (4,6)(4, 6). The second circle has center (3,0)(3, 0) and the same radius as the first. What is the equation of the line passing through the intersection points AA and BB?

  1. x+y−3=0x + y - 3 = 0
  2. x−y−1=0x - y - 1 = 0 (correct answer)
  3. x+y−1=0x + y - 1 = 0
  4. x−y+1=0x - y + 1 = 0
Explanation: First, find the radius: r=(4−1)2+(6−2)2=9+16=5r = \sqrt{(4-1)^2 + (6-2)^2} = \sqrt{9 + 16} = 5. Circle 1: (x−1)2+(y−2)2=25(x-1)^2 + (y-2)^2 = 25. Circle 2: (x−3)2+(y−0)2=25(x-3)^2 + (y-0)^2 = 25. The radical axis (line through intersection points) is found by subtracting the equations: (x−1)2+(y−2)2−[(x−3)2+y2]=0(x-1)^2 + (y-2)^2 - [(x-3)^2 + y^2] = 0. Expanding: x2−2x+1+y2−4y+4−x2+6x−9−y2=0x^2 - 2x + 1 + y^2 - 4y + 4 - x^2 + 6x - 9 - y^2 = 0. Simplifying: 4x−4y−4=04x - 4y - 4 = 0, or x−y−1=0x - y - 1 = 0. Choice A has wrong coefficients. Choice C has wrong constant term. Choice D has wrong sign on constant term.

Question 5

A circle has the equation (x−2)2+(y+1)2=9(x - 2)^2 + (y + 1)^2 = 9. If this circle is translated so that its center moves to the origin, and then dilated by a factor of 23\frac{2}{3} centered at the origin, what is the equation of the resulting circle?

  1. x2+y2=49x^2 + y^2 = \frac{4}{9}
  2. x2+y2=6x^2 + y^2 = 6
  3. x2+y2=9x^2 + y^2 = 9
  4. x2+y2=4x^2 + y^2 = 4 (correct answer)
Explanation: When you encounter circle transformations, you need to apply each transformation step-by-step to see how it affects the equation. The original circle (x−2)2+(y+1)2=9(x - 2)^2 + (y + 1)^2 = 9 has center (2,−1)(2, -1) and radius r=9=3r = \sqrt{9} = 3. First, translating the center to the origin means shifting the circle so its center moves from (2,−1)(2, -1) to (0,0)(0, 0). This eliminates the horizontal and vertical shifts in the equation, giving us x2+y2=9x^2 + y^2 = 9. The radius remains 3. Next, dilating by a factor of 23\frac{2}{3} centered at the origin multiplies all distances from the origin by 23\frac{2}{3}. Since the radius represents distance from the center, the new radius becomes 3×23=23 \times \frac{2}{3} = 2. Therefore, the final equation is x2+y2=22=4x^2 + y^2 = 2^2 = 4. Looking at the wrong answers: Choice A gives x2+y2=49x^2 + y^2 = \frac{4}{9}, which would result from incorrectly squaring the dilation factor (23)2\left(\frac{2}{3}\right)^2 and multiplying it by the original radius instead of the equation's constant term. Choice B gives x2+y2=6x^2 + y^2 = 6, which might come from adding the original radius and new radius (3+3×23=53 + 3 \times \frac{2}{3} = 5, though this doesn't quite match). Choice C gives x2+y2=9x^2 + y^2 = 9, which forgets to apply the dilation entirely. Remember: when dilating a circle, the radius gets multiplied by the dilation factor, so r2r^2 in the equation gets multiplied by the square of that factor.

Question 6

A communication tower is located at (3,4)(3,4). Its signal reaches all points within a radius of 1010 miles. What is the equation of the circle modeling the boundary of the coverage area?

  1. (x−4)2+(y−3)2=100(x-4)^2+(y-3)^2=100
  2. (x−3)2+(y−4)2=10(x-3)^2+(y-4)^2=10
  3. (x+3)2+(y+4)2=100(x+3)^2+(y+4)^2=100
  4. (x−3)2+(y−4)2=100(x-3)^2+(y-4)^2=100 (correct answer)
Explanation: This question tests your understanding of how to derive circle equations using the Pythagorean Theorem (distance formula) or find a circle's center and radius by completing the square. A circle is defined as all points at a fixed distance (radius r) from a center point (h, k): using the distance formula, the distance from any point (x, y) on the circle to the center is √[(x - h)² + (y - k)²] = r. Squaring both sides eliminates the radical and gives the standard form (x - h)² + (y - k)² = r². This equation comes directly from the Pythagorean Theorem applied to the right triangle formed by the horizontal distance (x - h), vertical distance (y - k), and radius r as hypotenuse! For the tower at (3, 4) with radius 10, it's (x - 3)² + (y - 4)² = 100, which is choice C. Choice A has the signs flipped, incorrectly centering at (-3, -4). Deriving from center and radius: (1) Write the distance from general point (x, y) to center (h, k) using distance formula: √[(x - h)² + (y - k)²], (2) Set equal to radius r, (3) Square both sides to get (x - h)² + (y - k)² = r². That's it! For example, center (2, -5) and radius 6: (x - 2)² + (y + 5)² = 36. The Pythagorean Theorem makes circle equations!

Question 7

A radio tower is located at (3,4)(3,4) on a coordinate grid. Its signal reaches all points within 10 units of the tower. Write the equation of the circle representing all points exactly 10 units from the tower.

  1. (x−4)2+(y−3)2=100(x-4)^2 + (y-3)^2 = 100
  2. (x−3)2+(y−4)2=100(x-3)^2 + (y-4)^2 = 100 (correct answer)
  3. (x−3)2+(y−4)2=10(x-3)^2 + (y-4)^2 = 10
  4. (x+3)2+(y+4)2=100(x+3)^2 + (y+4)^2 = 100
Explanation: This question tests your understanding of how to derive circle equations using the Pythagorean Theorem (distance formula) or find a circle's center and radius by completing the square. A circle is defined as all points at a fixed distance (radius r) from a center point (h, k): using the distance formula, the distance from any point (x, y) on the circle to the center is √[(x - h)² + (y - k)²] = r. Squaring both sides eliminates the radical and gives the standard form (x - h)² + (y - k)² = r². This equation comes directly from the Pythagorean Theorem applied to the right triangle formed by the horizontal distance (x - h), vertical distance (y - k), and radius r as hypotenuse! For a tower at (3, 4) with reach 10 units, it's (x - 3)² + (y - 4)² = 100. Choice C correctly derives the equation as (x-3)² + (y-4)² = 100. Something like choice A may forget to square the radius or mix up coordinates. Deriving from center and radius: (1) Write the distance from general point (x, y) to center (h, k) using distance formula: √[(x - h)² + (y - k)²], (2) Set equal to radius r, (3) Square both sides to get (x - h)² + (y - k)² = r². That's it! For example, center (2, -5) and radius 6: (x - 2)² + (y - (-5))² = 6², which simplifies to (x - 2)² + (y + 5)² = 36. The Pythagorean Theorem makes circle equations! Awesome application to real-world scenarios—keep it up!

Question 8

Find the center and radius of the circle x2+y2+10x−4y+13=0x^2 + y^2 + 10x - 4y + 13 = 0.

  1. Center (−5,2)(-5,2), radius 22
  2. Center (−10,4)(-10,4), radius 44
  3. Center (5,−2)(5,-2), radius 44
  4. Center (−5,2)(-5,2), radius 44 (correct answer)
Explanation: This question tests your understanding of how to derive circle equations using the Pythagorean Theorem (distance formula) or find a circle's center and radius by completing the square. When a circle equation is in general form x² + y² + Dx + Ey + F = 0, we complete the square in both x and y to reveal the center and radius: group x-terms (x² + Dx) and complete the square by adding (D/2)², do the same for y-terms with (E/2)², then rearrange to get (x + D/2)² + (y + E/2)² = (D/2)² + (E/2)² - F. The center is (-D/2, -E/2) and radius is √[(D/2)² + (E/2)² - F]. It's completing the square twice, once for each variable! For x² + y² + 10x - 4y + 13 = 0, completing the square gives (x + 5)² + (y - 2)² = 16, so center (-5, 2) and radius 4. Choice B correctly identifies the center and radius as (-5, 2) and 4. An error like in choice A might come from not flipping the signs for the center—always use center (-D/2, -E/2). Reading center from standard form has a sign trap: in (x - h)² + (y - k)² = r², the center is (h, k), but the signs in the equation are OPPOSITE! From (x - 3)² + (y + 2)² = 16, the center is (3, -2) because (x - 3) has center x = 3, and (y + 2) = (y - (-2)) has center y = -2. Think: what values make each squared term equal zero? Those are your center coordinates. Don't just copy the numbers—flip the signs! Great job verifying by completing the square—you've got this!

Question 9

Complete the square to write the circle equation x2+y2−8x+6y+9=0x^2+y^2-8x+6y+9=0 in standard form. Which standard-form equation is correct?

  1. (x−4)2+(y+3)2=16(x-4)^2+(y+3)^2=16 (correct answer)
  2. (x+4)2+(y−3)2=16(x+4)^2+(y-3)^2=16
  3. (x−4)2+(y+3)2=4(x-4)^2+(y+3)^2=4
  4. (x−8)2+(y+6)2=9(x-8)^2+(y+6)^2=9
Explanation: This question tests your understanding of how to derive circle equations using the Pythagorean Theorem (distance formula) or find a circle's center and radius by completing the square. When a circle equation is in general form x² + y² + Dx + Ey + F = 0, we complete the square in both x and y to reveal the center and radius: group x-terms (x² + Dx) and complete the square by adding (D/2)², do the same for y-terms with (E/2)², then rearrange to get (x + D/2)² + (y + E/2)² = (D/2)² + (E/2)² - F. The center is (-D/2, -E/2) and radius is √[(D/2)² + (E/2)² - F]. It's completing the square twice, once for each variable! Starting with x² + y² - 8x + 6y + 9 = 0, move 9 to get x² - 8x + y² + 6y = -9, complete square for x: (x - 4)² - 16, for y: (y + 3)² - 9, so (x - 4)² + (y + 3)² = -9 + 16 + 9 = 16. Choice A correctly completes the square to identify the standard form (x - 4)² + (y + 3)² = 16. A distractor like choice B might forget the signs when completing the square, using +4 for x instead of -4, but gently remember D = -8 so -D/2 = 4 for the center's x-coordinate. Completing the square for circles: (1) Group x-terms together and y-terms together: (x² + Dx) + (y² + Ey) = -F, (2) Complete square in x by adding (D/2)² to both sides, (3) Complete square in y by adding (E/2)² to both sides, (4) Factor: (x + D/2)² + (y + E/2)² = (D/2)² + (E/2)² - F, (5) Read center as (-D/2, -E/2) and radius as √[right side].

Question 10

Rewrite the circle x2+y2−2x+12y+20=0x^2+y^2-2x+12y+20=0 in standard form (x−h)2+(y−k)2=r2(x-h)^2+(y-k)^2=r^2 by completing the square.

  1. (x−2)2+(y+12)2=17(x-2)^2+(y+12)^2=17
  2. (x−1)2+(y+6)2=20(x-1)^2+(y+6)^2=20
  3. (x−1)2+(y+6)2=17(x-1)^2+(y+6)^2=17 (correct answer)
  4. (x+1)2+(y−6)2=17(x+1)^2+(y-6)^2=17
Explanation: This question tests your understanding of how to derive circle equations using the Pythagorean Theorem (distance formula) or find a circle's center and radius by completing the square. When a circle equation is in general form x² + y² + Dx + Ey + F = 0, we complete the square in both x and y to reveal the center and radius: group x-terms (x² + Dx) and complete the square by adding (D/2)², do the same for y-terms with (E/2)², then rearrange to get (x + D/2)² + (y + E/2)² = (D/2)² + (E/2)² - F. The center is (-D/2, -E/2) and radius is √[(D/2)² + (E/2)² - F]. It's completing the square twice, once for each variable! For x² + y² - 2x + 12y + 20 = 0, group and complete: (x² - 2x) + (y² + 12y) = -20, add 1 and 36 to both sides for (x - 1)² + (y + 6)² = 17. Choice A correctly completes the square to (x - 1)² + (y + 6)² = 17. Choice B flips the signs incorrectly, changing the center. Completing the square for circles: (1) Group x-terms together and y-terms together: (x² + Dx) + (y² + Ey) = -F, (2) Complete square in x by adding (D/2)² to both sides, (3) Complete square in y by adding (E/2)² to both sides, (4) Factor: (x + D/2)² + (y + E/2)² = (D/2)² + (E/2)² - F, (5) Read center as (-D/2, -E/2) and radius as √[right side]. Example: x² + y² + 6x - 8y = 0 → (x + 3)² + (y - 4)² = 25, so center (-3, 4), radius 5.

Question 11

Find the center and radius of the circle x2+y2+6x−4y−12=0x^2+y^2+6x-4y-12=0 by completing the square.

  1. Center (3,−2)(3,-2), radius 55
  2. Center (−3,2)(-3,2), radius 11
  3. Center (−3,2)(-3,2), radius 55 (correct answer)
  4. Center (−6,4)(-6,4), radius 55
Explanation: This question tests your understanding of how to derive circle equations using the Pythagorean Theorem (distance formula) or find a circle's center and radius by completing the square. When a circle equation is in general form x² + y² + Dx + Ey + F = 0, we complete the square in both x and y to reveal the center and radius: group x-terms (x² + Dx) and complete the square by adding (D/2)², do the same for y-terms with (E/2)², then rearrange to get (x + D/2)² + (y + E/2)² = (D/2)² + (E/2)² - F. The center is (-D/2, -E/2) and radius is √[(D/2)² + (E/2)² - F]. It's completing the square twice, once for each variable! For x² + y² + 6x - 4y - 12 = 0, group and complete: (x² + 6x) + (y² - 4y) = 12, add 9 and 4 to both sides for (x + 3)² + (y - 2)² = 25, so center (-3, 2) and radius 5. Choice A correctly identifies the center and radius as (-3, 2) and 5. Choice B flips the signs on the center, perhaps by not negating D/2 and E/2 properly. Completing the square for circles: (1) Group x-terms together and y-terms together: (x² + Dx) + (y² + Ey) = -F, (2) Complete square in x by adding (D/2)² to both sides, (3) Complete square in y by adding (E/2)² to both sides, (4) Factor: (x + D/2)² + (y + E/2)² = (D/2)² + (E/2)² - F, (5) Read center as (-D/2, -E/2) and radius as √[right side]. Example: x² + y² + 6x - 8y = 0 → (x + 3)² + (y - 4)² = 25, so center (-3, 4), radius 5.

Question 12

A circle has center at (3,−2)(3, -2) and passes through the point (7,1)(7, 1). If this circle is reflected across the line y=xy = x, what is the equation of the reflected circle?

  1. (x+2)2+(y−3)2=25(x + 2)^2 + (y - 3)^2 = 25 (correct answer)
  2. (x−2)2+(y+3)2=25(x - 2)^2 + (y + 3)^2 = 25
  3. (x+3)2+(y−2)2=25(x + 3)^2 + (y - 2)^2 = 25
  4. (x−3)2+(y+2)2=25(x - 3)^2 + (y + 2)^2 = 25
Explanation: First, find the radius using the distance formula: r=(7−3)2+(1−(−2))2=16+9=5r = \sqrt{(7-3)^2 + (1-(-2))^2} = \sqrt{16 + 9} = 5. The original circle has equation (x−3)2+(y+2)2=25(x-3)^2 + (y+2)^2 = 25. When reflecting across y=xy = x, coordinates (a,b)(a,b) become (b,a)(b,a). So the center (3,−2)(3,-2) becomes (−2,3)(-2,3), giving equation (x+2)2+(y−3)2=25(x+2)^2 + (y-3)^2 = 25. Choice B reflects the signs incorrectly. Choice C uses the original center. Choice D is the original equation.

Question 13

Use the Pythagorean Theorem idea (distance from center) to write the equation of the circle with center (0,0)(0,0) passing through the point (5,12)(5,12).

  1. (x−5)2+(y−12)2=169(x-5)^2+(y-12)^2=169
  2. x2+y2=169x^2+y^2=169 (correct answer)
  3. x2+y2=13x^2+y^2=13
  4. x2+y2=25x^2+y^2=25
Explanation: This question tests your understanding of how to derive circle equations using the Pythagorean Theorem (distance formula) or find a circle's center and radius by completing the square. A circle is defined as all points at a fixed distance (radius r) from a center point (h, k): using the distance formula, the distance from any point (x, y) on the circle to the center is √[(x - h)² + (y - k)²] = r. Squaring both sides eliminates the radical and gives the standard form (x - h)² + (y - k)² = r². This equation comes directly from the Pythagorean Theorem applied to the right triangle formed by the horizontal distance (x - h), vertical distance (y - k), and radius r as hypotenuse! For center (0,0) passing through (5,12), the radius is √(5² + 12²) = √169 = 13, so x² + y² = 169, choice B. Choice A uses 13 without squaring it for r², a common mistake. Deriving from center and radius: (1) Write the distance from general point (x, y) to center (h, k) using distance formula: √[(x - h)² + (y - k)²], (2) Set equal to radius r, (3) Square both sides to get (x - h)² + (y - k)² = r². That's it! For example, center (2, -5) and radius 6: (x - 2)² + (y + 5)² = 36. The Pythagorean Theorem makes circle equations!

Question 14

Derive the equation of a circle with center (0,0)(0,0) and radius 33 using the idea that the distance from (x,y)(x,y) to the center is 33.

  1. x2+y2=6x^2 + y^2 = 6
  2. (x−3)2+(y−3)2=9(x-3)^2 + (y-3)^2 = 9
  3. x2+y2=9x^2 + y^2 = 9 (correct answer)
  4. x2+y2=3x^2 + y^2 = 3
Explanation: This question tests your understanding of how to derive circle equations using the Pythagorean Theorem (distance formula) or find a circle's center and radius by completing the square. A circle is defined as all points at a fixed distance (radius r) from a center point (h, k): using the distance formula, the distance from any point (x, y) on the circle to the center is (x−h)2+(y−k)2=r\sqrt{(x - h)^2 + (y - k)^2} = r. Squaring both sides eliminates the radical and gives the standard form (x−h)2+(y−k)2=r2(x - h)^2 + (y - k)^2 = r^2. This equation comes directly from the Pythagorean Theorem applied to the right triangle formed by the horizontal distance (x - h), vertical distance (y - k), and radius r as hypotenuse! For center (0, 0) and radius 3, it's simply x2+y2=9x^2 + y^2 = 9, since (x−0)2+(y−0)2=32(x - 0)^2 + (y - 0)^2 = 3^2. Choice C correctly derives the equation with r² = 9 for the origin-centered circle. A distractor like choice B might use 6 instead of 9, perhaps confusing radius with diameter, but remember to square the radius. Deriving from center and radius: (1) Write the distance from general point (x, y) to center (h, k) using distance formula: (x−h)2+(y−k)2\sqrt{(x - h)^2 + (y - k)^2}, (2) Set equal to radius r, (3) Square both sides to get (x−h)2+(y−k)2=r2(x - h)^2 + (y - k)^2 = r^2. That's it!

Question 15

Complete the square: x2+y2+4x−10y+13=0x^2 + y^2 + 4x - 10y + 13 = 0. Which standard form equation is correct?

  1. (x+2)2+(y−5)2=16(x+2)^2 + (y-5)^2 = 16 (correct answer)
  2. (x−2)2+(y+5)2=16(x-2)^2 + (y+5)^2 = 16
  3. (x+2)2+(y−5)2=4(x+2)^2 + (y-5)^2 = 4
  4. (x+4)2+(y−10)2=16(x+4)^2 + (y-10)^2 = 16
Explanation: This question tests your understanding of how to derive circle equations using the Pythagorean Theorem (distance formula) or find a circle's center and radius by completing the square. When a circle equation is in general form x² + y² + Dx + Ey + F = 0, we complete the square in both x and y to reveal the center and radius: group x-terms (x² + Dx) and complete the square by adding (D/2)², do the same for y-terms with (E/2)², then rearrange to get (x + D/2)² + (y + E/2)² = (D/2)² + (E/2)² - F. The center is (-D/2, -E/2) and radius is √[(D/2)² + (E/2)² - F]. It's completing the square twice, once for each variable! For x² + y² + 4x - 10y + 13 = 0, it becomes (x + 2)² + (y - 5)² = 16. Choice A correctly completes the square to (x+2)² + (y-5)² = 16. A common issue, as in choice B, is mishandling signs when rewriting the center. Completing the square for circles: (1) Group x-terms together and y-terms together: (x² + Dx) + (y² + Ey) = -F, (2) Complete square in x by adding (D/2)² to both sides, (3) Complete square in y by adding (E/2)² to both sides, (4) Factor: (x + D/2)² + (y + E/2)² = (D/2)² + (E/2)² - F, (5) Read center as (-D/2, -E/2) and radius as √[right side]. Example: x² + y² + 6x - 8y = 0 → (x + 3)² + (y - 4)² = 9 + 16 = 25, so center (-3, 4), radius 5. Super progress—you're a pro at this now!

Question 16

Rewrite the general-form circle equation x2+y2+10x−2y+17=0x^2+y^2+10x-2y+17=0 in standard form (x−h)2+(y−k)2=r2(x-h)^2+(y-k)^2=r^2.

  1. (x−5)2+(y+1)2=9(x-5)^2+(y+1)^2=9
  2. (x+5)2+(y−1)2=9(x+5)^2+(y-1)^2=9 (correct answer)
  3. (x+5)2+(y−1)2=25(x+5)^2+(y-1)^2=25
  4. (x+10)2+(y−2)2=17(x+10)^2+(y-2)^2=17
Explanation: This question tests your understanding of how to derive circle equations using the Pythagorean Theorem (distance formula) or find a circle's center and radius by completing the square. When a circle equation is in general form x² + y² + Dx + Ey + F = 0, we complete the square in both x and y to reveal the center and radius: group x-terms (x² + Dx) and complete the square by adding (D/2)², do the same for y-terms with (E/2)², then rearrange to get (x + D/2)² + (y + E/2)² = (D/2)² + (E/2)² - F. The center is (-D/2, -E/2) and radius is √[(D/2)² + (E/2)² - F]. It's completing the square twice, once for each variable! For x² + y² + 10x - 2y + 17 = 0, move 17: x² + 10x + y² - 2y = -17, complete x: (x + 5)² - 25, y: (y - 1)² - 1, so (x + 5)² + (y - 1)² = -17 + 25 + 1 = 9. Choice A correctly rewrites to standard form with center (-5, 1) and r² = 9. A common error, like in choice D, is miscalculating the right side to 25 instead of 9, but gently check the arithmetic: -17 + 25 + 1 is indeed 9. Completing the square for circles: (1) Group x-terms together and y-terms together: (x² + Dx) + (y² + Ey) = -F, (2) Complete square in x by adding (D/2)² to both sides, (3) Complete square in y by adding (E/2)² to both sides, (4) Factor: (x + D/2)² + (y + E/2)² = (D/2)² + (E/2)² - F, (5) Read center as (-D/2, -E/2) and radius as √[right side]. Example: x² + y² + 6x - 8y = 0 → (x + 3)² + (y - 4)² = 9 + 16 = 25, so center (-3, 4), radius 5.

Question 17

Find the center and radius of the circle given by x2+y2+6x−4y+9=0x^2+y^2+6x-4y+9=0.

  1. Center (−3,2)(-3,2), radius 44
  2. Center (3,−2)(3,-2), radius 22
  3. Center (−3,2)(-3,2), radius 22 (correct answer)
  4. Center (−6,4)(-6,4), radius 22
Explanation: This question tests your understanding of how to derive circle equations using the Pythagorean Theorem (distance formula) or find a circle's center and radius by completing the square. When a circle equation is in general form x² + y² + Dx + Ey + F = 0, we complete the square in both x and y to reveal the center and radius: group x-terms (x² + Dx) and complete the square by adding (D/2)², do the same for y-terms with (E/2)², then rearrange to get (x + D/2)² + (y + E/2)² = (D/2)² + (E/2)² - F. The center is (-D/2, -E/2) and radius is √[(D/2)² + (E/2)² - F]. It's completing the square twice, once for each variable! For x² + y² + 6x - 4y + 9 = 0, move 9: x² + 6x + y² - 4y = -9, complete x: (x + 3)² - 9, y: (y - 2)² - 4, so (x + 3)² + (y - 2)² = -9 + 9 + 4 = 4, center (-3, 2), radius 2. Choice B correctly identifies the center (-3, 2) and radius 2 after completing the square. A mistake like in choice A might not flip the signs for the center, using (3, -2) instead, but remember center is (-D/2, -E/2). Reading center from standard form has a sign trap: in (x - h)² + (y - k)² = r², the center is (h, k), but the signs in the equation are OPPOSITE! From (x - 3)² + (y + 2)² = 16, the center is (3, -2) because (x - 3) has center x = 3, and (y + 2) = (y - (-2)) has center y = -2. Think: what values make each squared term equal zero? Those are your center coordinates. Don't just copy the numbers—flip the signs!

Question 18

What is the equation of a circle centered at (−2,5)(-2,5) with radius 77? (Use the distance formula idea that every point on the circle is 77 units from the center.)

  1. (x−2)2+(y+5)2=49(x-2)^2+(y+5)^2=49
  2. (x+2)2+(y−5)2=7(x+2)^2+(y-5)^2=7
  3. (x+2)2+(y−5)2=49(x+2)^2+(y-5)^2=49 (correct answer)
  4. (x−2)2+(y−5)2=49(x-2)^2+(y-5)^2=49
Explanation: This question tests your understanding of how to derive circle equations using the Pythagorean Theorem (distance formula) or find a circle's center and radius by completing the square. A circle is defined as all points at a fixed distance (radius r) from a center point (h, k): using the distance formula, the distance from any point (x, y) on the circle to the center is √[(x - h)² + (y - k)²] = r. Squaring both sides eliminates the radical and gives the standard form (x - h)² + (y - k)² = r². This equation comes directly from the Pythagorean Theorem applied to the right triangle formed by the horizontal distance (x - h), vertical distance (y - k), and radius r as hypotenuse! With center (-2, 5) and radius 7, we substitute h = -2, k = 5, and r = 7 into the standard form: (x - (-2))² + (y - 5)² = 7², which simplifies to (x + 2)² + (y - 5)² = 49. Choice C correctly derives the equation as (x + 2)² + (y - 5)² = 49. Choice B incorrectly has r² = 7 instead of r² = 49, forgetting to square the radius. Deriving from center and radius: (1) Write the distance from general point (x, y) to center (h, k) using distance formula: √[(x - h)² + (y - k)²], (2) Set equal to radius r, (3) Square both sides to get (x - h)² + (y - k)² = r². That's it! For example, center (2, -5) and radius 6: (x - 2)² + (y - (-5))² = 6², which simplifies to (x - 2)² + (y + 5)² = 36. The Pythagorean Theorem makes circle equations!

Question 19

Complete the square to rewrite x2+y2+10x−2y+17=0x^2+y^2+10x-2y+17=0 in standard form (x−h)2+(y−k)2=r2(x-h)^2+(y-k)^2=r^2.

  1. (x+5)2+(y−1)2=9(x+5)^2+(y-1)^2=9 (correct answer)
  2. (x−5)2+(y+1)2=9(x-5)^2+(y+1)^2=9
  3. (x+10)2+(y−2)2=17(x+10)^2+(y-2)^2=17
  4. (x+5)2+(y−1)2=17(x+5)^2+(y-1)^2=17
Explanation: This question tests your understanding of how to derive circle equations using the Pythagorean Theorem (distance formula) or find a circle's center and radius by completing the square. When a circle equation is in general form x² + y² + Dx + Ey + F = 0, we complete the square in both x and y to reveal the center and radius: group x-terms (x² + Dx) and complete the square by adding (D/2)², do the same for y-terms with (E/2)², then rearrange to get (x + D/2)² + (y + E/2)² = (D/2)² + (E/2)² - F. The center is (-D/2, -E/2) and radius is √[(D/2)² + (E/2)² - F]. It's completing the square twice, once for each variable! For x² + y² + 10x - 2y + 17 = 0, group and complete: (x² + 10x) + (y² - 2y) = -17, add 25 and 1 to both sides for (x + 5)² + (y - 1)² = 9. Choice A correctly completes the square to (x + 5)² + (y - 1)² = 9. Choice B flips the signs incorrectly, which would change the center to (5, -1) instead of watching the completing process. Completing the square for circles: (1) Group x-terms together and y-terms together: (x² + Dx) + (y² + Ey) = -F, (2) Complete square in x by adding (D/2)² to both sides, (3) Complete square in y by adding (E/2)² to both sides, (4) Factor: (x + D/2)² + (y + E/2)² = (D/2)² + (E/2)² - F, (5) Read center as (-D/2, -E/2) and radius as √[right side]. Example: x² + y² + 6x - 8y = 0 → (x + 3)² + (y - 4)² = 25, so center (-3, 4), radius 5.

Question 20

Use the Pythagorean Theorem (distance formula) to derive the equation of the circle with center (2,−3)(2,-3) and radius 44. What is the equation in standard form (x−h)2+(y−k)2=r2(x-h)^2+(y-k)^2=r^2?

  1. (x−2)2+(y−3)2=16(x-2)^2+(y-3)^2=16
  2. (x−2)2+(y+3)2=4(x-2)^2+(y+3)^2=4
  3. (x+2)2+(y−3)2=16(x+2)^2+(y-3)^2=16
  4. (x−2)2+(y+3)2=16(x-2)^2+(y+3)^2=16 (correct answer)
Explanation: This question tests your understanding of how to derive circle equations using the Pythagorean Theorem (distance formula) or find a circle's center and radius by completing the square. A circle is defined as all points at a fixed distance (radius r) from a center point (h, k): using the distance formula, the distance from any point (x, y) on the circle to the center is √[(x - h)² + (y - k)²] = r. Squaring both sides eliminates the radical and gives the standard form (x - h)² + (y - k)² = r². This equation comes directly from the Pythagorean Theorem applied to the right triangle formed by the horizontal distance (x - h), vertical distance (y - k), and radius r as hypotenuse! For the center at (2, -3) and radius 4, plug in to get (x - 2)² + (y - (-3))² = 4², which simplifies to (x - 2)² + (y + 3)² = 16. Choice B correctly derives the equation with the proper signs for the center and squared radius. A common mistake, like in choice D, is flipping the sign for y, resulting in (y - 3) instead of (y + 3), but remember to use (y - k) where k = -3 means +3. Deriving from center and radius: (1) Write the distance from general point (x, y) to center (h, k) using distance formula: √[(x - h)² + (y - k)²], (2) Set equal to radius r, (3) Square both sides to get (x - h)² + (y - k)² = r². That's it! Reading center from standard form has a sign trap: in (x - h)² + (y - k)² = r², the center is (h, k), but the signs in the equation are OPPOSITE!