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Geometry Help: Derive The Equation Of A Circle

Review real example questions for Derive The Equation Of A Circle in Geometry.

Question 1 / 10

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Which expression represents all points (x,y)(x,y) that are a fixed distance 5 from the center (0,−3)(0,-3)?

Use the distance formula to represent "distance from (x,y)(x,y) to (0,−3)(0,-3) equals 5."

All questions

Question 1

Which expression represents all points (x,y)(x,y) that are a fixed distance 5 from the center (0,−3)(0,-3)?

Use the distance formula to represent "distance from (x,y)(x,y) to (0,−3)(0,-3) equals 5."

  1. x2+(y+3)2=5x^2+(y+3)^2=5
  2. x2+(y−3)2=25x^2+(y-3)^2=25
  3. x2+(y+3)2=25x^2+(y+3)^2=25 (correct answer)
  4. x2+y2=25x^2+y^2=25

Explanation: The skill here is deriving the equation of a circle. A circle is defined as the set of all points in a plane that are at a fixed distance, called the radius, from a fixed point, called the center. Apply the distance formula to set distance from (x,y) to (0,-3) equal to 5. This produces the form x^2 + (y + 3)^2 = 25, linking center and radius algebraically. This justifies choice C as representing the fixed distance of 5. A distractor is reversing the center's y-sign, as in choice B, changing it to (0,3). Always think about the geometric distance before algebra to confirm the equation's structure.

Question 2

Which expression represents all points (x,y)(x,y) that are a fixed distance 66 from the center (−1,3)(-1,3)?

  1. (x+1)2+(y−3)2=6(x+1)^2+(y-3)^2=6
  2. (x+1)2+(y−3)2=36(x+1)^2+(y-3)^2=36 (correct answer)
  3. (x−1)2+(y+3)2=36(x-1)^2+(y+3)^2=36
  4. x2+y2=36x^2+y^2=36

Explanation: This question directly translates the geometric definition of a circle into its algebraic equation. A circle is the set of all points at a fixed distance from a center point. For points (x,y) that are distance 6 from center (-1,3), the distance formula gives √[(x-(-1))² + (y-3)²] = 6. Squaring both sides yields (x+1)² + (y-3)² = 36. Notice that the radius 6 becomes 36 when squared—this is crucial for the standard form. A common error is writing (x+1)² + (y-3)² = 6, forgetting that the equation uses r², not r. Remember: the equation of a circle always involves the square of the radius, reflecting the squared terms in the distance formula.

Question 3

A circle is the set of all points a fixed distance from its center. What is the equation of the circle with center (2,−5)(2, -5) and radius 33?​

  1. (x−2)2+(y+5)2=9(x-2)^2+(y+5)^2=9 (correct answer)
  2. (x+2)2+(y−5)2=9(x+2)^2+(y-5)^2=9
  3. (x−2)2+(y+5)2=3(x-2)^2+(y+5)^2=3
  4. x2+y2=9x^2+y^2=9

Explanation: This question directly applies the definition of a circle to write its equation. A circle is the set of all points that are equidistant from a fixed center point, creating a perfect round shape. Using the distance formula, every point (x, y) on a circle with center (2, -5) and radius 3 satisfies √[(x - 2)² + (y - (-5))²] = 3. Squaring both sides to eliminate the square root gives us (x - 2)² + (y + 5)² = 9, which is the standard form of the circle equation. Notice how the center coordinates appear in the parentheses with opposite signs: center (2, -5) gives us (x - 2) and (y + 5). A common mistake is writing (y - 5) instead of (y + 5) when the y-coordinate is negative—always be careful with signs. To master circle equations, remember that they encode the simple geometric idea that every point maintains the same distance from the center.

Question 4

A circle is the set of all points a fixed distance from its center. What is the equation of the circle with center (−2,3)(-2, 3) and radius 44?​

  1. (x+2)2+(y−3)2=16(x+2)^2+(y-3)^2=16 (correct answer)
  2. (x−2)2+(y+3)2=16(x-2)^2+(y+3)^2=16
  3. (x+2)2+(y−3)2=4(x+2)^2+(y-3)^2=4
  4. x2+y2=16x^2+y^2=16

Explanation: This question asks you to derive the equation of a circle from its geometric properties. A circle is defined as the set of all points that are equidistant from a fixed center point. Using the distance formula, any point (x, y) on the circle must satisfy √[(x - h)² + (y - k)²] = r, where (h, k) is the center and r is the radius. Squaring both sides gives us the standard form: (x - h)² + (y - k)² = r². For a circle with center (-2, 3) and radius 4, we substitute to get (x - (-2))² + (y - 3)² = 4², which simplifies to (x + 2)² + (y - 3)² = 16. A common mistake is confusing the signs when the center has negative coordinates—remember that (x - h) becomes (x + 2) when h = -2. To avoid errors, always think of the distance relationship first before manipulating the algebra.

Question 5

What is the equation of the circle with center (−2,−1)(-2,-1) and radius 33? (Justify using that all points on the circle are the same distance from the center.)

  1. (x+2)2+(y+1)2=3(x+2)^2+(y+1)^2=3
  2. (x−2)2+(y−1)2=9(x-2)^2+(y-1)^2=9
  3. (x+2)2+(y+1)2=9(x+2)^2+(y+1)^2=9 (correct answer)
  4. x2+y2=9x^2+y^2=9

Explanation: To write a circle's equation given its center and radius, we apply the geometric definition directly. A circle is the set of all points that are the same distance from a fixed center point. If the center is (-2, -1) and radius is 3, then any point (x, y) on the circle satisfies the distance formula: √[(x-(-2))² + (y-(-1))²] = 3. Squaring both sides gives (x+2)² + (y+1)² = 9. The standard form (x-h)² + (y-k)² = r² makes the center and radius immediately visible. A common error involves the signs: since we have center (-2, -1), the equation uses (x-(-2)) = (x+2) and (y-(-1)) = (y+1). When given center and radius, think of the distance relationship first, then translate to algebra.

Question 6

A circle has the equation x2+y2−6x+8y−11=0x^2 + y^2 - 6x + 8y - 11 = 0. What is the equation of the circle that has the same radius but is centered at the origin?

  1. x2+y2=36x^2 + y^2 = 36 (correct answer)
  2. x2+y2=6x^2 + y^2 = 6
  3. x2+y2=64x^2 + y^2 = 64
  4. x2+y2=100x^2 + y^2 = 100

Explanation: Complete the square for the given equation: x2−6x+y2+8y=11x^2 - 6x + y^2 + 8y = 11. For x: x2−6x=(x−3)2−9x^2 - 6x = (x-3)^2 - 9. For y: y2+8y=(y+4)2−16y^2 + 8y = (y+4)^2 - 16. Substituting: (x−3)2−9+(y+4)2−16=11(x-3)^2 - 9 + (y+4)^2 - 16 = 11, which gives (x−3)2+(y+4)2=36(x-3)^2 + (y+4)^2 = 36. The radius is 36=6\sqrt{36} = 6. A circle centered at the origin with the same radius has equation x2+y2=36x^2 + y^2 = 36. Choice B uses the radius instead of radius squared. Choices C and D are incorrect calculations of the radius squared.

Question 7

Which statement correctly identifies the center and radius of the circle given by the equation x2+y2−6x+4y−12=0x^2+y^2-6x+4y-12=0? (Use completing the square to interpret the geometry.)

  1. Center (3,−2)(3,-2) and radius 55 (correct answer)
  2. Center (−3,2)(-3,2) and radius 55
  3. Center (3,−2)(3,-2) and radius 2525
  4. Center (0,0)(0,0) and radius 55

Explanation: This problem requires deriving the center and radius from the general form of a circle equation. A circle is defined as all points equidistant from its center. To interpret x² + y² - 6x + 4y - 12 = 0, we complete the square for both variables to reach standard form (x-h)² + (y-k)² = r². For x terms: x² - 6x = (x-3)² - 9; for y terms: y² + 4y = (y+2)² - 4. Substituting gives (x-3)² + (y+2)² - 9 - 4 - 12 = 0, which simplifies to (x-3)² + (y+2)² = 25. This reveals center (3, -2) and radius √25 = 5. Students often mistake r² for r, thinking the radius is 25 instead of 5. Remember: completing the square transforms the distance relationship into readable geometric parameters.

Question 8

Which expression represents all points (x,y)(x,y) that are a fixed distance 55 from the center (−3,4)(-3,4)? (Use the distance formula.)

  1. (x+3)2+(y−4)2=5(x+3)^2+(y-4)^2=5
  2. (x+3)2+(y−4)2=25(x+3)^2+(y-4)^2=25 (correct answer)
  3. (x−3)2+(y+4)2=25(x-3)^2+(y+4)^2=25
  4. x2+y2=25x^2+y^2=25

Explanation: This problem directly applies the definition of a circle to derive its equation. A circle is the set of all points at a constant distance from a center point. Using the distance formula, if (x, y) is distance 5 from center (-3, 4), then √[(x-(-3))² + (y-4)²] = 5. Squaring both sides gives (x+3)² + (y-4)² = 25. The standard form (x-h)² + (y-k)² = r² clearly shows center (h, k) = (-3, 4) and radius² = 25. Students often confuse r with r², writing 5 instead of 25 on the right side, which would incorrectly suggest radius = √5. When translating from geometric description to algebraic equation, always square the given radius to match the squared distance formula.

Question 9

Which statement correctly identifies the center and radius of the circle given by x2+y2+4x−10y+13=0x^2 + y^2 + 4x - 10y + 13 = 0? Rewrite the equation in standard form to interpret it geometrically.

  1. Center (2,−5)(2,-5) and radius 44
  2. Center (−2,5)(-2,5) and radius 44 (correct answer)
  3. Center (−2,5)(-2,5) and radius 1616
  4. Center (0,0)(0,0) and radius 44

Explanation: The skill here is deriving the equation of a circle. A circle is defined as the set of all points in a plane that are at a fixed distance, called the radius, from a fixed point, called the center. The distance formula applies implicitly when completing the square on the given equation. This reveals the form (x+2)2+(y−5)2=16(x + 2)^2 + (y - 5)^2 = 16, connecting to center (−2,5)(-2,5) and radius 44. This justifies choice B as the geometric interpretation. A misconception is failing to square the half-coefficients correctly, leading to wrong radius like 1616 instead of 44 in choice C. Always think about the geometric distance before algebra to verify the completed form.

Question 10

Which equation follows from the geometric definition of a circle with center (−2,1)(-2,1) that passes through the point (1,5)(1,5)?

Use the idea that every point on the circle is the same distance from the center.

  1. (x+2)2+(y−1)2=25(x+2)^2+(y-1)^2=25 (correct answer)
  2. (x−2)2+(y+1)2=25(x-2)^2+(y+1)^2=25
  3. (x+2)2+(y−1)2=5(x+2)^2+(y-1)^2=5
  4. x2+y2=25x^2+y^2=25

Explanation: The skill here is deriving the equation of a circle from its center and a point it passes through. A circle is defined as the set of all points in a plane that are equidistant from a fixed point called the center. The distance formula applies by calculating the radius as the distance from the center (-2, 1) to (1, 5), which is (1+2)2+(5−1)2=9+16=5\sqrt{(1 + 2)^2 + (5 - 1)^2} = \sqrt{9 + 16} = 5. This connects algebraically to (x+2)2+(y−1)2=25(x + 2)^2 + (y - 1)^2 = 25, the standard form with center (-2, 1) and r2=25r^2 = 25. Option A is justified because it matches this derivation from the geometric definition. A distractor like option C uses r=5r = 5 without squaring to 25, misunderstanding the equation form. To transfer, always compute the distance to find the radius before forming the equation.