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Algebra 2 Flashcards: Derive The Equation Of A Circle

Study Derive The Equation Of A Circle in Algebra 2 with focused flashcards that help you recognize the idea, recall the key rule, and apply it in practice-style prompts.

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What this deck covers

This deck focuses on Derive The Equation Of A Circle, giving you a quick way to review the definitions, rules, and examples that matter most for Algebra 2.

How to use these flashcards

Work through these flashcards in short sessions. Try to answer each prompt before flipping the card, then revisit any cards you miss until the explanation feels automatic.

Algebra 2 Flashcards: Derive The Equation Of A Circle

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QUESTION

What midpoint formula gives the center of a circle from diameter endpoints (x1,y1)(x_1,y_1)(x1​,y1​) and (x2,y2)(x_2,y_2)(x2​,y2​)?

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ANSWER

(x1+x22,y1+y22)\left(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2}\right)(2x1​+x2​​,2y1​+y2​​). Midpoint formula gives center from endpoints.

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Flashcard 1: What midpoint formula gives the center of a circle from diameter endpoints (x1,y1)(x_1,y_1)(x1​,y1​) and (x2,y2)(x_2,y_2)(x2​,y2​)?

Answer: (x1+x22,y1+y22)\left(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2}\right)(2x1​+x2​​,2y1​+y2​​). Midpoint formula gives center from endpoints.

Flashcard 2: What is the radius in terms of diameter endpoints (x1,y1)(x_1,y_1)(x1​,y1​) and (x2,y2)(x_2,y_2)(x2​,y2​)?

Answer: r=12(x2−x1)2+(y2−y1)2r=\frac{1}{2}\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}r=21​(x2​−x1​)2+(y2​−y1​)2​. Radius is half the distance between endpoints.

Flashcard 3: What is the general form after expanding x2+(y−4)2=20x^2+(y-4)^2=20x2+(y−4)2=20?

Answer: x2+y2−8y−4=0x^2+y^2-8y-4=0x2+y2−8y−4=0. Expand and rearrange to general form.

Flashcard 4: Identify the value added to complete the square for x2−14xx^2-14xx2−14x.

Answer: 494949. Take half of coefficient: (−14/2)2=49(-14/2)^2=49(−14/2)2=49.

Flashcard 5: What is the completed-square expression equivalent to y2+8yy^2+8yy2+8y?

Answer: (y+4)2−16(y+4)^2-16(y+4)2−16. Complete square: (y+4)2=y2+8y+16(y+4)^2=y^2+8y+16(y+4)2=y2+8y+16.

Flashcard 6: What is the center and radius of (x+1)2+(y−2)2=0(x+1)^2+(y-2)^2=0(x+1)2+(y−2)2=0?

Answer: Center (−1,2)(-1,2)(−1,2); radius 000. Degenerate circle: single point at (−1,2)(-1,2)(−1,2).

Flashcard 7: What is the center and radius of (x−7)2+(y+3)2=81(x-7)^2+(y+3)^2=81(x−7)2+(y+3)2=81?

Answer: Center (7,−3)(7,-3)(7,−3); radius 999. Read center and radius from standard form.

Flashcard 8: What is the center and radius of x2+y2−12x−2y+37=0x^2+y^2-12x-2y+37=0x2+y2−12x−2y+37=0?

Answer: Center (6,1)(6,1)(6,1); radius 000. Complete square: r2=36+1−37=0r^2=36+1-37=0r2=36+1−37=0.

Flashcard 9: What is the center of the circle x2+y2−2x+14y+33=0x^2+y^2-2x+14y+33=0x2+y2−2x+14y+33=0?

Answer: (1,−7)(1,-7)(1,−7). From standard form, center is (1,−7)(1,-7)(1,−7).

Flashcard 10: What is the general form of a circle equation after expanding standard form?

Answer: x2+y2+Dx+Ey+F=0x^2+y^2+Dx+Ey+F=0x2+y2+Dx+Ey+F=0. Result of expanding standard form and collecting terms.

Flashcard 11: What is the general form after expanding (x+5)2+(y−1)2=36(x+5)^2+(y-1)^2=36(x+5)2+(y−1)2=36?

Answer: x2+y2+10x−2y−10=0x^2+y^2+10x-2y-10=0x2+y2+10x−2y−10=0. Expand and rearrange to general form.

Flashcard 12: What is the equation of the circle with diameter endpoints (0,0)(0,0)(0,0) and (6,8)(6,8)(6,8)?

Answer: (x−3)2+(y−4)2=25(x-3)^2+(y-4)^2=25(x−3)2+(y−4)2=25. Midpoint (3,4)(3,4)(3,4), radius half diagonal length.

Flashcard 13: What is the center of the circle with diameter endpoints (2,−1)(2,-1)(2,−1) and (8,5)(8,5)(8,5)?

Answer: (5,2)(5,2)(5,2). Midpoint formula for diameter endpoints.

Flashcard 14: What is the radius of the circle with diameter endpoints (2,−1)(2,-1)(2,−1) and (8,5)(8,5)(8,5)?

Answer: 323\sqrt{2}32​. Half the distance between diameter endpoints.

Flashcard 15: What is the equation of the circle with diameter endpoints (−2,4)(-2,4)(−2,4) and (4,−2)(4,-2)(4,−2)?

Answer: (x−1)2+(y−1)2=18(x-1)^2+(y-1)^2=18(x−1)2+(y−1)2=18. Midpoint (1,1)(1,1)(1,1), radius half diagonal length.

Flashcard 16: What midpoint formula gives the center of a circle from diameter endpoints (x1,y1)(x_1,y_1)(x1​,y1​) and (x2,y2)(x_2,y_2)(x2​,y2​)?

Answer: (x1+x22,y1+y22)\left(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2}\right)(2x1​+x2​​,2y1​+y2​​). Midpoint formula gives center from endpoints.

Flashcard 17: What is the radius in terms of diameter endpoints (x1,y1)(x_1,y_1)(x1​,y1​) and (x2,y2)(x_2,y_2)(x2​,y2​)?

Answer: r=12(x2−x1)2+(y2−y1)2r=\frac{1}{2}\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}r=21​(x2​−x1​)2+(y2​−y1​)2​. Radius is half the distance between endpoints.

Flashcard 18: What is the equation of the circle in standard form: x2+y2=9x^2+y^2=9x2+y2=9?

Answer: (x−0)2+(y−0)2=9(x-0)^2+(y-0)^2=9(x−0)2+(y−0)2=9. Already in standard form with center (0,0)(0,0)(0,0).

Flashcard 19: What is the general form after expanding (x−2)2+(y+3)2=16(x-2)^2+(y+3)^2=16(x−2)2+(y+3)2=16?

Answer: x2+y2−4x+6y−3=0x^2+y^2-4x+6y-3=0x2+y2−4x+6y−3=0. Expand and rearrange to general form.

Flashcard 20: What is the general form after expanding (x+5)2+(y−1)2=36(x+5)^2+(y-1)^2=36(x+5)2+(y−1)2=36?

Answer: x2+y2+10x−2y−10=0x^2+y^2+10x-2y-10=0x2+y2+10x−2y−10=0. Expand and rearrange to general form.

Flashcard 21: What is the general form after expanding x2+(y−4)2=20x^2+(y-4)^2=20x2+(y−4)2=20?

Answer: x2+y2−8y−4=0x^2+y^2-8y-4=0x2+y2−8y−4=0. Expand and rearrange to general form.

Flashcard 22: Identify the value added to complete the square for x2−14xx^2-14xx2−14x.

Answer: 494949. Take half of coefficient: (−14/2)2=49(-14/2)^2=49(−14/2)2=49.

Flashcard 23: Identify the value added to complete the square for y2+9yy^2+9yy2+9y.

Answer: 814\frac{81}{4}481​. Take half of coefficient: (9/2)2=81/4(9/2)^2=81/4(9/2)2=81/4.

Flashcard 24: What is the completed-square expression equivalent to x2−12xx^2-12xx2−12x?

Answer: (x−6)2−36(x-6)^2-36(x−6)2−36. Complete square: (x−6)2=x2−12x+36(x-6)^2=x^2-12x+36(x−6)2=x2−12x+36.

Flashcard 25: What is the completed-square expression equivalent to y2+8yy^2+8yy2+8y?

Answer: (y+4)2−16(y+4)^2-16(y+4)2−16. Complete square: (y+4)2=y2+8y+16(y+4)^2=y^2+8y+16(y+4)2=y2+8y+16.

Flashcard 26: What is the center and radius of (x+1)2+(y−2)2=0(x+1)^2+(y-2)^2=0(x+1)2+(y−2)2=0?

Answer: Center (−1,2)(-1,2)(−1,2); radius 000. Degenerate circle: single point at (−1,2)(-1,2)(−1,2).

Flashcard 27: What is the center and radius of (x−7)2+(y+3)2=81(x-7)^2+(y+3)^2=81(x−7)2+(y+3)2=81?

Answer: Center (7,−3)(7,-3)(7,−3); radius 999. Read center and radius from standard form.

Flashcard 28: What is the standard form equation of a circle with center (h,k)(h,k)(h,k) and radius rrr?

Answer: (x−h)2+(y−k)2=r2(x-h)^2+(y-k)^2=r^2(x−h)2+(y−k)2=r2. Standard form shows center (h,k)(h,k)(h,k) and radius rrr explicitly.

Flashcard 29: What is the center and radius of the circle (x−h)2+(y−k)2=r2(x-h)^2+(y-k)^2=r^2(x−h)2+(y−k)2=r2?

Answer: Center (h,k)(h,k)(h,k); radius rrr. Read center and radius directly from standard form.

Flashcard 30: What is the equation of a circle centered at the origin with radius rrr?

Answer: x2+y2=r2x^2+y^2=r^2x2+y2=r2. When center is at origin, h=0h=0h=0 and k=0k=0k=0.