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Geometry

Geometry Help: Derive The Equation Of A Circle

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

Question 1

Which equation follows from the geometric definition of a circle for the set of all points (x,y)(x,y)(x,y) that are a distance 666 from the center (1,−3)(1,-3)(1,−3)?

  1. (x−1)2+(y+3)2=6(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(x−1)2+(y+3)2=36
  3. (x+1)2+(y−3)2=36(x+1)^2+(y-3)^2=36(x+1)2+(y−3)2=36
  4. x2+y2=36x^2+y^2=36x2+y2=36
Explanation: To derive the equation of a circle, we start with its fundamental definition. A circle consists of all points that maintain a constant distance from a fixed center point. Using the distance formula, if a point (x, y) is distance 6 from center (1, -3), then √[(x-1)² + (y-(-3))²] = 6. Squaring both sides eliminates the radical: (x-1)² + (y+3)² = 36. This standard form clearly shows the center at (1, -3) and radius² = 36, confirming radius = 6. A common mistake is writing r instead of r² on the right side, leading to equation (x-1)² + (y+3)² = 6. When deriving circle equations, always square the radius value to match the squared terms on the left.

Question 2

The equation x2+y2−8x+6y+9=0x^2 + y^2 - 8x + 6y + 9 = 0x2+y2−8x+6y+9=0 represents a circle. After completing the square, what are the coordinates of the center and the radius?

  1. Center: (4,−3)(4, -3)(4,−3), Radius: 444
  2. Center: (4,−3)(4, -3)(4,−3), Radius: 222
  3. Center: (−4,3)(-4, 3)(−4,3), Radius: 444
  4. Center: (8,−6)(8, -6)(8,−6), Radius: 16\sqrt{16}16​
Explanation: Complete the square for both variables. For x terms: x2−8x=(x−4)2−16x^2 - 8x = (x-4)^2 - 16x2−8x=(x−4)2−16. For y terms: y2+6y=(y+3)2−9y^2 + 6y = (y+3)^2 - 9y2+6y=(y+3)2−9. Substituting: (x−4)2−16+(y+3)2−9+9=0(x-4)^2 - 16 + (y+3)^2 - 9 + 9 = 0(x−4)2−16+(y+3)2−9+9=0, which simplifies to (x−4)2+(y+3)2=16(x-4)^2 + (y+3)^2 = 16(x−4)2+(y+3)2=16. The center is (4,−3)(4, -3)(4,−3) and radius is 16=4\sqrt{16} = 416​=4. Choice B has the correct center but wrong radius. Choice C has sign errors in the center coordinates. Choice D uses the original coefficients incorrectly.

Question 3

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

  1. (x−3)2+(y−1)2=25(x - 3)^2 + (y - 1)^2 = 25(x−3)2+(y−1)2=25
  2. (x−4)2+(y−1)2=25(x - 4)^2 + (y - 1)^2 = 25(x−4)2+(y−1)2=25
  3. (x−3)2+(y−1)2=100(x - 3)^2 + (y - 1)^2 = 100(x−3)2+(y−1)2=100
  4. (x−3)2+(y+1)2=25(x - 3)^2 + (y + 1)^2 = 25(x−3)2+(y+1)2=25
Explanation: The center of the circle is the midpoint of the diameter: (−1+72,4+(−2)2)=(3,1)\left(\frac{-1+7}{2}, \frac{4+(-2)}{2}\right) = (3, 1)(2−1+7​,24+(−2)​)=(3,1). The radius is half the distance between A and B: r=12(7−(−1))2+(−2−4)2=1264+36=12100=5r = \frac{1}{2}\sqrt{(7-(-1))^2 + (-2-4)^2} = \frac{1}{2}\sqrt{64+36} = \frac{1}{2}\sqrt{100} = 5r=21​(7−(−1))2+(−2−4)2​=21​64+36​=21​100​=5. Therefore r2=25r^2 = 25r2=25. The equation is (x−3)2+(y−1)2=25(x-3)^2 + (y-1)^2 = 25(x−3)2+(y−1)2=25. Choice B has an incorrect x-coordinate for the center. Choice C uses the diameter squared instead of radius squared. Choice D has an incorrect y-coordinate for the center.

Question 4

Which statement correctly identifies the center and radius of the circle given by x2+y2+2x−12y+29=0x^2+y^2+2x-12y+29=0x2+y2+2x−12y+29=0?

  1. Center (−1,6)(-1,6)(−1,6), radius 8\sqrt{8}8​
  2. Center (1,−6)(1,-6)(1,−6), radius 8\sqrt{8}8​
  3. Center (−1,6)(-1,6)(−1,6), radius 888
  4. Center (0,0)(0,0)(0,0), radius 29\sqrt{29}29​
Explanation: The skill involves deriving the equation of a circle using its geometric properties. A circle is defined as the set of all points in a plane that are a fixed distance, called the radius, from a fixed point, called the center. This definition directly translates to the distance formula, where the distance between any point (x,y) on the circle and the center (h,k) equals the radius r. Algebraically, this becomes (x - h)^2 + (y - k)^2 = r^2, with completing the square revealing these elements. For the given equation, completing the square yields center (-1,6) and radius √8, matching the geometric interpretation. A common distractor misconception is failing to square root for the radius, leading to 8 instead of √8. To transfer this strategy, always think about the distance from the center before jumping into algebraic manipulations.

Question 5

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)(2,−5) and radius 333?​

  1. (x−2)2+(y+5)2=9(x-2)^2+(y+5)^2=9(x−2)2+(y+5)2=9
  2. (x+2)2+(y−5)2=9(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(x−2)2+(y+5)2=3
  4. x2+y2=9x^2+y^2=9x2+y2=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 6

What is the equation of the circle with center (−2,−1)(-2,-1)(−2,−1) and radius 333? (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(x+2)2+(y+1)2=3
  2. (x−2)2+(y−1)2=9(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(x+2)2+(y+1)2=9
  4. x2+y2=9x^2+y^2=9x2+y2=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 7

A circle has the equation x2+y2−6x+8y−11=0x^2 + y^2 - 6x + 8y - 11 = 0x2+y2−6x+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 = 36x2+y2=36
  2. x2+y2=6x^2 + y^2 = 6x2+y2=6
  3. x2+y2=64x^2 + y^2 = 64x2+y2=64
  4. x2+y2=100x^2 + y^2 = 100x2+y2=100
Explanation: Complete the square for the given equation: x2−6x+y2+8y=11x^2 - 6x + y^2 + 8y = 11x2−6x+y2+8y=11. For x: x2−6x=(x−3)2−9x^2 - 6x = (x-3)^2 - 9x2−6x=(x−3)2−9. For y: y2+8y=(y+4)2−16y^2 + 8y = (y+4)^2 - 16y2+8y=(y+4)2−16. Substituting: (x−3)2−9+(y+4)2−16=11(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(x−3)2+(y+4)2=36. The radius is 36=6\sqrt{36} = 636​=6. A circle centered at the origin with the same radius has equation x2+y2=36x^2 + y^2 = 36x2+y2=36. Choice B uses the radius instead of radius squared. Choices C and D are incorrect calculations of the radius squared.

Question 8

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=0x2+y2−6x+4y−12=0? (Use completing the square to interpret the geometry.)

  1. Center (3,−2)(3,-2)(3,−2) and radius 555
  2. Center (−3,2)(-3,2)(−3,2) and radius 555
  3. Center (3,−2)(3,-2)(3,−2) and radius 252525
  4. Center (0,0)(0,0)(0,0) and radius 555
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 9

Which statement correctly identifies the center and radius of the circle given by x2+y2−6x+4y=12x^2+y^2-6x+4y=12x2+y2−6x+4y=12? (Interpret by completing the square to reveal the geometric meaning.)

  1. Center (3,−2)(3,-2)(3,−2), radius 555
  2. Center (−3,2)(-3,2)(−3,2), radius 555
  3. Center (3,−2)(3,-2)(3,−2), radius 252525
  4. Center (0,0)(0,0)(0,0), radius 12\sqrt{12}12​
Explanation: The skill involves deriving the equation of a circle using its geometric properties. A circle is defined as the set of all points in a plane that are a fixed distance, called the radius, from a fixed point, called the center. This definition directly translates to the distance formula, where the distance between any point (x,y) on the circle and the center (h,k) equals the radius r. Algebraically, this becomes (x - h)^2 + (y - k)^2 = r^2, and completing the square reveals the center and radius from the general form. For the given equation, completing the square yields center (3,-2) and radius 5, matching the geometric interpretation. A common distractor misconception is mistaking the radius for its square, leading to radius 25 instead of 5. To transfer this strategy, always think about the distance from the center before jumping into algebraic manipulations.

Question 10

Which statement correctly identifies the center and radius of the circle given by x2+y2+8x−2y−8=0x^2+y^2+8x-2y-8=0x2+y2+8x−2y−8=0?

  1. Center (−4,1)(-4,1)(−4,1) and radius 555
  2. Center (4,−1)(4,-1)(4,−1) and radius 555
  3. Center (−4,1)(-4,1)(−4,1) and radius 252525
  4. Center (0,0)(0,0)(0,0) and radius 555
Explanation: This question requires completing the square to find a circle's center and radius from general form. A circle is defined as all points at a fixed distance from a center. Starting with x² + y² + 8x - 2y - 8 = 0, we rearrange and complete the square: (x² + 8x + 16) + (y² - 2y + 1) = 8 + 16 + 1. This gives (x+4)² + (y-1)² = 25, revealing center (-4,1) and radius √25 = 5. The transformation from general to standard form exposes the geometric meaning: the center coordinates are the opposites of what appears inside the parentheses. Students often forget that completing the square for (x + 8x) requires adding 16, not 64. Think of completing the square as revealing the hidden distance relationship.