Derive the Equation of a Circle - Geometry

1. The endpoints of a diameter of a circle are $A(-1, 4)$ and $B(7, -2)$. Which equation represents this circle?
   Answer: $(x - 3)^2 + (y - 1)^2 = 25$ The center of the circle is the midpoint of the diameter: $\left(\frac{-1+7}{2}, \frac{4+(-2)}{2}\right) = (3, 1)$. The radius is half the distance between A and B: $r = \frac{1}{2}\sqrt{(7-(-1))^2 + (-2-4)^2} = \frac{1}{2}\sqrt{64+36} = \frac{1}{2}\sqrt{100} = 5$. Therefore $r^2 = 25$. The equation is $(x-3)^2 + (y-1)^2 = 25$. 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.
2. Refer to the figure below. A circle is drawn on the coordinate plane with a horizontal radius shown from the center to a point on the circle. Which equation represents this circle?
   Answer: $(x-3)^2 + (y-2)^2 = 16$ The center is at $(3,2)$ and the radius extends to $(7,2)$, giving $r=4$ and $r^2=16$. Choice B uses $r$ instead of $r^2$. Choice C reverses the signs of the center. Choice D confuses the endpoint of the radius with the center.
3. Which statement correctly identifies the center and radius of the circle given by the equation $x^2+y^2+8x-2y+13=0$?
   Answer: Center $(-4,1)$ and radius $2$ To find the center and radius from x² + y² + 8x - 2y + 13 = 0, we must complete the square. A circle is defined as all points at a fixed distance from its center. Rearranging: (x² + 8x) + (y² - 2y) = -13. For x: x² + 8x = (x+4)² - 16; for y: y² - 2y = (y-1)² - 1. Substituting yields (x+4)² + (y-1)² - 16 - 1 = -13, which simplifies to (x+4)² + (y-1)² = 4. This reveals center (-4, 1) and radius √4 = 2. A common error is misidentifying the center's signs: since we have (x+4)², the center's x-coordinate is -4, not 4. When completing the square, remember that (x-h)² means the center is at positive h, while (x+h)² means the center is at negative h.
4. Which statement correctly identifies the center and radius of the circle given by $x^2+y^2+2x-12y+29=0$?
   Answer: Center $(-1,6)$, radius $\sqrt{8}$ 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.
5. Which statement correctly identifies the center and radius of the circle given by $x^2+y^2+8x-2y-8=0$?
   Answer: Center $(-4,1)$ and radius $5$ 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.
6. The equation $x^2 + y^2 - 8x + 6y + 9 = 0$ represents a circle. After completing the square, what are the coordinates of the center and the radius?
   Answer: Center: $(4, -3)$, Radius: $4$ Complete the square for both variables. For x terms: $x^2 - 8x = (x-4)^2 - 16$. For y terms: $y^2 + 6y = (y+3)^2 - 9$. Substituting: $(x-4)^2 - 16 + (y+3)^2 - 9 + 9 = 0$, which simplifies to $(x-4)^2 + (y+3)^2 = 16$. The center is $(4, -3)$ and radius is $\sqrt{16} = 4$. 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.
7. Which statement correctly identifies the center and radius of the circle given by $x^2+y^2-6x+4y-12=0$?
   Answer: Center $(3,-2)$ and radius $5$ This question requires converting a circle equation from general form to standard form to identify its center and radius. A circle is defined as all points equidistant from a center point. To find the center and radius from x² + y² - 6x + 4y - 12 = 0, we complete the square for both x and y terms. Grouping and completing: (x² - 6x + 9) + (y² + 4y + 4) = 12 + 9 + 4, which gives (x-3)² + (y+2)² = 25. This reveals the center at (3,-2) and radius √25 = 5. A common mistake is confusing radius with r²: the right side equals r², not r. Think geometrically: completing the square reveals the distance relationship hidden in the expanded form.
8. What is the equation of the circle with center $(5,0)$ and radius $6$? Your equation should describe all points that are 6 units from $(5,0)$ on the coordinate plane.
   Answer: $(x-5)^2+y^2=36$ 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 equate distance from (x,y) to (5,0) to 6. This forms $(x - 5)^2$ + $y^2$ = 36, tying algebra to center and radius. This justifies choice C as describing points 6 units from (5,0). A misconception is neglecting to square the radius, as in choice A with 6 instead of 36. Always think about the geometric distance before algebra to build the proper equation.
9. Which statement correctly identifies the center and radius of the circle given by $x^2-6x+y^2+4y=12?$ Your reasoning should connect the equation to the set of points a fixed distance from a center.
   Answer: Center $ (3,-2) $ and radius $\sqrt{25}$ The skill here is deriving the equation of a circle and identifying its center and radius from a given equation. 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 setting the distance from any point (x, y) on the circle to the center (h, k) equal to the radius r, leading to $\sqrt{(x - h)^2 + (y - k)^2} = r$. Connecting this to algebra, completing the square rewrites the equation into $ $(x - h)^2$ + $(y - k)^2$ = $r^2$ $, revealing the center and radius. Option A is justified because completing the square on $ $x^2$ - 6x + $y^2$ + 4y = 12 $ gives $ $(x - 3)^2$ + $(y + 2)^2$ = 25 $, so center $ (3, -2) $ and radius $\sqrt{25} = 5$. A distractor like option C mistakes $ $r^2$ = 25 $ for r = 25, forgetting to take the square root. To transfer this, always prioritize the distance interpretation before algebraic steps like completing the square.
10. Which expression represents all points $(x,y)$ a fixed distance $7$ from the center $(-3,4)$?
    Answer: $(x+3)^2+(y-4)^2=49$ This question directly applies the geometric definition of a circle to derive its equation. A circle is defined as all points that are a fixed distance from a center point. Given that all points (x, y) must be exactly 7 units from center (-3, 4), we use the distance formula: √[(x - (-3))² + (y - 4)²] = 7. Squaring both sides to eliminate the radical gives (x + 3)² + (y - 4)² = 49. The equation form (x - h)² + (y - k)² = r² transparently shows the center at (h, k) = (-3, 4) and radius r = 7. Notice how the center coordinate -3 becomes +3 in the equation due to the subtraction: x - (-3) = x + 3. A common mistake is writing the radius as 7 instead of 49 on the right side, forgetting that we squared the entire equation. Always remember that the distance formula squared yields r², not r.
11. A circle is the set of all points a fixed distance from its center. What is the equation of the circle with center $(2, -5)$ and radius $3$?​
    Answer: $(x-2)^2+(y+5)^2=9$ 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.
12. What is the equation of the circle with center $(2,3)$ and radius $5$? Choose the equation that describes all points $(x,y)$ whose distance from $(2,3)$ is 5.
    Answer: $(x-2)^2+(y-3)^2=25$ The skill here is deriving the equation of a circle given its center and radius directly. 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 sets $\sqrt{(x - 2)^2 + (y - 3)^2} = 5$ for center $(2, 3)$ and radius 5. This connects algebraically to $(x - 2)^2 + (y - 3)^2 = 25$ by squaring both sides. Option A is justified as it precisely describes points at distance 5 from $(2, 3)$. A distractor like option C forgets to square the radius, using 5 instead of 25. To transfer, conceptualize the constant distance before writing the squared equation.
13. Which statement correctly identifies the center and radius of the circle given by the equation $x^2+y^2-6x+4y-12=0$? (Use completing the square to interpret the geometry.)
    Answer: Center $(3,-2)$ and radius $5$ 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.
14. A circle is the set of all points $(x,y)$ that are 6 units from the center $(0, -3)$. Which expression represents all points a fixed distance from the center?​
    Answer: $(x-0)^2+(y+3)^2=36$ This question directly applies the geometric definition of a circle to write its equation. A circle consists of all points that are exactly the same distance from a fixed center point. Using the distance formula, any point (x, y) that is 6 units from center (0, -3) must satisfy √[(x - 0)² + (y - (-3))²] = 6. Squaring both sides to eliminate the radical gives us (x - 0)² + (y + 3)² = 36, which simplifies to x² + (y + 3)² = 36. Notice that the radius 6 becomes 36 when squared—this is a crucial detail in the standard form equation. A common mistake is writing the radius instead of radius squared on the right side, which would give the incorrect equation x² + (y + 3)² = 6. Always remember that the standard form uses r², not r, because it comes from squaring the distance equation.
15. Which equation follows from the geometric definition of a circle with center $(1,-2)$ and radius $3$? Use the idea that the distance from any point $(x,y)$ on the circle to the center is 3.
    Answer: $(x-1)^2+(y+2)^2=9$ 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 directly by setting the distance from (x,y) to (1,-2) equal to 3. This yields the algebraic form $(x - 1)^2$ + $(y + 2)^2$ = 9, connecting center and radius. This justifies choice C as following from the geometric definition. A misconception is using the radius without squaring, as in choice A where 3 replaces 9. Always think about the geometric distance before algebra to ensure the equation matches the definition.
16. Which equation follows from the geometric definition of a circle with center $(-2,-4)$ and radius $5$ (all points a distance 5 from the center)?
    Answer: $(x+2)^2+(y+4)^2=25$ This question directly applies the definition of a circle to write its equation. A circle is the locus of all points at a constant distance from a center point. Using the distance formula, any point (x,y) that is distance 5 from center (-2,-4) satisfies √[(x-(-2))² + (y-(-4))²] = 5. Squaring eliminates the radical: (x+2)² + (y+4)² = 25. The standard form reveals the geometric relationship: we measure squared distance from the center. A common error is writing (x-2)² + (y-4)² = 25, which would incorrectly place the center at (2,4). Remember: in the standard form (x-h)² + (y-k)², the center is at (h,k), so negative center coordinates become positive in the equation.
17. Which expression represents all points $(x,y)$ that are 6 units from the center $(1,-3)$? (Use the distance formula connection.)
    Answer: $(x-1)^2+(y+3)^2=36$ 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 $(x - h)^2$ + $(y - k)^2$ = $r^2$, linking the equation to center and radius. For points 6 units from (1,-3), the equation is $(x-1)^2$ + $(y+3)^2$ = 36, as the radius squared is 36. A common distractor misconception is reversing the center signs, resulting in $(x+1)^2$ + $(y-3)^2$ = 36. To transfer this strategy, always think about the distance from the center before jumping into algebraic manipulations.
18. What is the equation of the circle with center $(-2,-1)$ and radius $3$? (Justify using that all points on the circle are the same distance from the center.)
    Answer: $(x+2)^2+(y+1)^2=9$ 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.
19. Which equation follows from the geometric definition of a circle with center $(-2,1)$ that passes through the point $(1,5)$? Use the idea that every point on the circle is the same distance from the center.
    Answer: $(x+2)^2+(y-1)^2=25$ 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 $\sqrt{(1 + 2)^2 + (5 - 1)^2} = \sqrt{9 + 16} = 5$. This connects algebraically to $ $(x + 2)^2$ + $(y - 1)^2$ = 25 $, the standard form with center (-2, 1) and $ $r^2$ = 25 $. Option A is justified because it matches this derivation from the geometric definition. A distractor like option C uses $ r = 5 $ without squaring to 25, misunderstanding the equation form. To transfer, always compute the distance to find the radius before forming the equation.
20. Which statement correctly identifies the center and radius of the circle given by $x^2+4x+y^2-12y=9$?
    Answer: Center $(-2,6)$, radius $\sqrt{49}$ This problem involves transforming a circle equation to standard form through completing the square. A circle is defined as all points equidistant from a center point. Starting with x² + 4x + y² - 12y = 9, we complete the square for each variable: x² + 4x becomes (x + 2)² - 4, and y² - 12y becomes (y - 6)² - 36. Substituting these expressions gives (x + 2)² + (y - 6)² = 9 + 4 + 36 = 49. The standard form (x - h)² + (y - k)² = r² reveals center (h, k) = (-2, 6) and radius r = √49 = 7. Note that (x + 2)² corresponds to a center x-coordinate of -2, while (y - 6)² gives a y-coordinate of 6. A common error is stating the radius as 49 instead of √49 = 7, misunderstanding that r² appears in the equation. When completing the square, remember to add the same values to both sides to maintain the equation's validity.

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