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.

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.

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.

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.

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.

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.

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 by rewriting the given equation through completing the square to reveal the center and radius. This connects to the standard form $ (x - 3)^2 + (y + 4)^2 = 25 $, identifying center $ (3,-4) $ and radius $ 5 $. This justifies choice A as the correct interpretation after algebraic manipulation. A distractor is misinterpreting the completed square constants, leading to reversed signs like in choice B. Always think about the geometric distance before algebra to guide the completing the square process.

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.

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. To find the equation, we apply the distance formula to express that the distance from (x,y) to the center (-3,4) equals 2. This leads to the algebraic form (x + $3)^2$ + (y - $4)^2$ = 4, linking the center and radius directly. This justifies choice C as the correct equation representing all points at distance 2 from (-3,4). A distractor misconception is squaring the radius incorrectly, as in choice A where it's set to 2 instead of 4. Always think about the geometric distance before algebra to avoid sign or squaring errors.

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.