Derive the Equation of a Circle - Algebra 2

Complete the square to rewrite $x^2+y^2+10x-2y+17=0$ in standard form $(x-h)^2+(y-k)^2=r^2$ .
Answer: $(x+5)^2+(y-1)^2=9$ This question tests your understanding of how to derive circle equations using the Pythagorean Theorem (distance formula) or find a circle's center and radius by completing the square. When a circle equation is in general form x² + y² + Dx + Ey + F = 0, we complete the square in both x and y to reveal the center and radius: group x-terms (x² + Dx) and complete the square by adding (D/2)², do the same for y-terms with (E/2)², then rearrange to get (x + D/2)² + (y + E/2)² = (D/2)² + (E/2)² - F. The center is (-D/2, -E/2) and radius is √[(D/2)² + (E/2)² - F]. It's completing the square twice, once for each variable! For x² + y² + 10x - 2y + 17 = 0, group and complete: (x² + 10x) + (y² - 2y) = -17, add 25 and 1 to both sides for (x + 5)² + (y - 1)² = 9. Choice A correctly completes the square to (x + 5)² + (y - 1)² = 9. Choice B flips the signs incorrectly, which would change the center to (5, -1) instead of watching the completing process. Completing the square for circles: (1) Group x-terms together and y-terms together: (x² + Dx) + (y² + Ey) = -F, (2) Complete square in x by adding (D/2)² to both sides, (3) Complete square in y by adding (E/2)² to both sides, (4) Factor: (x + D/2)² + (y + E/2)² = (D/2)² + (E/2)² - F, (5) Read center as (-D/2, -E/2) and radius as √[right side]. Example: x² + y² + 6x - 8y = 0 → (x + 3)² + (y - 4)² = 25, so center (-3, 4), radius 5.
Derive the equation of a circle with center $(4, -1)$ and radius $3$ .
Answer: $(x-4)^2 + (y+1)^2 = 9$ This question tests your understanding of how to derive circle equations using the Pythagorean Theorem (distance formula) or find a circle's center and radius by completing the square. A circle is defined as all points at a fixed distance (radius r) from a center point (h, k): using the distance formula, the distance from any point (x, y) on the circle to the center is √[(x - h)² + (y - k)²] = r. Squaring both sides eliminates the radical and gives the standard form (x - h)² + (y - k)² = r². This equation comes directly from the Pythagorean Theorem applied to the right triangle formed by the horizontal distance (x - h), vertical distance (y - k), and radius r as hypotenuse! For center (4, -1) and radius 3, it's (x - 4)² + (y - (-1))² = 9, or (x - 4)² + (y + 1)² = 9. Choice A correctly derives the equation as (x-4)² + (y+1)² = 9. A mistake like in choice B could be from incorrect sign handling for the y-coordinate—remember to subtract k, so negative k becomes + in the equation. Deriving from center and radius: (1) Write the distance from general point (x, y) to center (h, k) using distance formula: √[(x - h)² + (y - k)²], (2) Set equal to radius r, (3) Square both sides to get (x - h)² + (y - k)² = r². That's it! For example, center (2, -5) and radius 6: (x - 2)² + (y - (-5))² = 6², which simplifies to (x - 2)² + (y + 5)² = 36. The Pythagorean Theorem makes circle equations! Excellent work deriving it step by step—keep going!
Use the Pythagorean Theorem idea (distance from center) to write the equation of the circle with center $(0,0)$ passing through the point $(5,12)$ .
Answer: $x^2+y^2=169$ This question tests your understanding of how to derive circle equations using the Pythagorean Theorem (distance formula) or find a circle's center and radius by completing the square. A circle is defined as all points at a fixed distance (radius r) from a center point (h, k): using the distance formula, the distance from any point (x, y) on the circle to the center is √[(x - h)² + (y - k)²] = r. Squaring both sides eliminates the radical and gives the standard form (x - h)² + (y - k)² = r². This equation comes directly from the Pythagorean Theorem applied to the right triangle formed by the horizontal distance (x - h), vertical distance (y - k), and radius r as hypotenuse! For center (0,0) passing through (5,12), the radius is √(5² + 12²) = √169 = 13, so x² + y² = 169, choice B. Choice A uses 13 without squaring it for r², a common mistake. Deriving from center and radius: (1) Write the distance from general point (x, y) to center (h, k) using distance formula: √[(x - h)² + (y - k)²], (2) Set equal to radius r, (3) Square both sides to get (x - h)² + (y - k)² = r². That's it! For example, center (2, -5) and radius 6: (x - 2)² + (y + 5)² = 36. The Pythagorean Theorem makes circle equations!
All points $(x,y)$ that are 5 units from the point $(1,2)$ form a circle. Write the equation of this circle in standard form. (Use the Pythagorean Theorem/distance formula.)
Answer: $(x-1)^2 + (y-2)^2 = 25$ This question tests your understanding of how to derive circle equations using the Pythagorean Theorem (distance formula) or find a circle's center and radius by completing the square. A circle is defined as all points at a fixed distance (radius r) from a center point (h, k): using the distance formula, the distance from any point (x, y) on the circle to the center is √[(x - h)² + (y - k)²] = r. Squaring both sides eliminates the radical and gives the standard form (x - h)² + (y - k)² = r². This equation comes directly from the Pythagorean Theorem applied to the right triangle formed by the horizontal distance (x - h), vertical distance (y - k), and radius r as hypotenuse! Here, points 5 units from (1, 2) satisfy (x - 1)² + (y - 2)² = 25. Choice C correctly derives the equation as (x-1)² + (y-2)² = 25. A distractor like choice A forgets to square the radius—it's r², not r! Deriving from center and radius: (1) Write the distance from general point (x, y) to center (h, k) using distance formula: √[(x - h)² + (y - k)²], (2) Set equal to radius r, (3) Square both sides to get (x - h)² + (y - k)² = r². That's it! For example, center (2, -5) and radius 6: (x - 2)² + (y - (-5))² = 6², which simplifies to (x - 2)² + (y + 5)² = 36. The Pythagorean Theorem makes circle equations! Fantastic— you're nailing the distance concept!
Rewrite the circle equation $x^2+y^2+6x-4y+9=0$ in standard form $(x-h)^2+(y-k)^2=r^2$ .
Answer: $(x+3)^2+(y-2)^2=4$ This question tests your understanding of how to derive circle equations using the Pythagorean Theorem (distance formula) or find a circle's center and radius by completing the square. When a circle equation is in general form x² + y² + Dx + Ey + F = 0, we complete the square in both x and y to reveal the center and radius: group x-terms (x² + Dx) and complete the square by adding (D/2)², do the same for y-terms with (E/2)², then rearrange to get (x + D/2)² + (y + E/2)² = (D/2)² + (E/2)² - F. The center is (-D/2, -E/2) and radius is √[(D/2)² + (E/2)² - F]. It's completing the square twice, once for each variable! Starting with x² + y² + 6x - 4y + 9 = 0, we rearrange: (x² + 6x) + (y² - 4y) = -9. Complete the square in x: add (6/2)² = 9 to both sides. Complete the square in y: add (-4/2)² = 4 to both sides. This gives (x + 3)² + (y - 2)² = -9 + 9 + 4 = 4. Choice A correctly shows (x + 3)² + (y - 2)² = 4. Choice D incorrectly has r² = 16 instead of r² = 4. Completing the square for circles: (1) Group x-terms together and y-terms together: (x² + Dx) + (y² + Ey) = -F, (2) Complete square in x by adding (D/2)² to both sides, (3) Complete square in y by adding (E/2)² to both sides, (4) Factor: (x + D/2)² + (y + E/2)² = (D/2)² + (E/2)² - F, (5) Read center as (-D/2, -E/2) and radius as √[right side]. Reading center from standard form has a sign trap: in (x - h)² + (y - k)² = r², the center is (h, k), but the signs in the equation are OPPOSITE!
What are the center and radius of the circle in standard form $(x-7)^2+(y+2)^2=49$ ?
Answer: Center $(7,-2)$ , radius $7$ This question tests your understanding of how to derive circle equations using the Pythagorean Theorem (distance formula) or find a circle's center and radius by completing the square. A circle is defined as all points at a fixed distance (radius r) from a center point (h, k): using the distance formula, the distance from any point (x, y) on the circle to the center is √[(x - h)² + (y - k)²] = r. Squaring both sides eliminates the radical and gives the standard form (x - h)² + (y - k)² = r². This equation comes directly from the Pythagorean Theorem applied to the right triangle formed by the horizontal distance (x - h), vertical distance (y - k), and radius r as hypotenuse! From (x - 7)² + (y + 2)² = 49, the center is (7, -2) and radius is 7, as in choice A. Choice B flips the signs without accounting for the opposites in the equation. Reading center from standard form has a sign trap: in (x - h)² + (y - k)² = r², the center is (h, k), but the signs in the equation are OPPOSITE! From (x - 3)² + (y + 2)² = 16, the center is (3, -2) because (x - 3) has center x = 3, and (y + 2) = (y - (-2)) has center y = -2. Think: what values make each squared term equal zero? Those are your center coordinates. Don't just copy the numbers—flip the signs!
A radio tower is located at $(3,4)$ on a coordinate grid. Its signal reaches all points within 10 units of the tower. Write the equation of the circle representing all points exactly 10 units from the tower.
Answer: $(x-3)^2 + (y-4)^2 = 100$ This question tests your understanding of how to derive circle equations using the Pythagorean Theorem (distance formula) or find a circle's center and radius by completing the square. A circle is defined as all points at a fixed distance (radius r) from a center point (h, k): using the distance formula, the distance from any point (x, y) on the circle to the center is √[(x - h)² + (y - k)²] = r. Squaring both sides eliminates the radical and gives the standard form (x - h)² + (y - k)² = r². This equation comes directly from the Pythagorean Theorem applied to the right triangle formed by the horizontal distance (x - h), vertical distance (y - k), and radius r as hypotenuse! For a tower at (3, 4) with reach 10 units, it's (x - 3)² + (y - 4)² = 100. Choice C correctly derives the equation as (x-3)² + (y-4)² = 100. Something like choice A may forget to square the radius or mix up coordinates. Deriving from center and radius: (1) Write the distance from general point (x, y) to center (h, k) using distance formula: √[(x - h)² + (y - k)²], (2) Set equal to radius r, (3) Square both sides to get (x - h)² + (y - k)² = r². That's it! For example, center (2, -5) and radius 6: (x - 2)² + (y - (-5))² = 6², which simplifies to (x - 2)² + (y + 5)² = 36. The Pythagorean Theorem makes circle equations! Awesome application to real-world scenarios—keep it up!
A circle has center at $(3, -2)$ and passes through the point $(7, 1)$ . If this circle is reflected across the line $y = x$ , what is the equation of the reflected circle?
Answer: $(x + 2)^2 + (y - 3)^2 = 25$ First, find the radius using the distance formula: $r = \sqrt{(7-3)^2 + (1-(-2))^2} = \sqrt{16 + 9} = 5$ . The original circle has equation $(x-3)^2 + (y+2)^2 = 25$ . When reflecting across $y = x$ , coordinates $(a,b)$ become $(b,a)$ . So the center $(3,-2)$ becomes $(-2,3)$ , giving equation $(x+2)^2 + (y-3)^2 = 25$ . Choice B reflects the signs incorrectly. Choice C uses the original center. Choice D is the original equation.
All points $(x,y)$ that are $5$ units from the point $(1,2)$ form a circle. Using the distance formula, write the equation of the circle in standard form.
Answer: $(x-1)^2+(y-2)^2=25$ This question tests your understanding of how to derive circle equations using the Pythagorean Theorem (distance formula) or find a circle's center and radius by completing the square. A circle is defined as all points at a fixed distance (radius r) from a center point (h, k): using the distance formula, the distance from any point (x, y) on the circle to the center is √[(x - h)² + (y - k)²] = r. Squaring both sides eliminates the radical and gives the standard form (x - h)² + (y - k)² = r². This equation comes directly from the Pythagorean Theorem applied to the right triangle formed by the horizontal distance (x - h), vertical distance (y - k), and radius r as hypotenuse! All points 5 units from (1, 2) means center (1, 2) and radius 5. Using the distance formula: √[(x - 1)² + (y - 2)²] = 5. Squaring both sides: (x - 1)² + (y - 2)² = 25. Choice C correctly derives the equation as (x - 1)² + (y - 2)² = 25. Choice A incorrectly has r² = 10 instead of r² = 25, forgetting to square the radius 5. Deriving from center and radius: (1) Write the distance from general point (x, y) to center (h, k) using distance formula: √[(x - h)² + (y - k)²], (2) Set equal to radius r, (3) Square both sides to get (x - h)² + (y - k)² = r². That's it! For example, center (2, -5) and radius 6: (x - 2)² + (y - (-5))² = 6², which simplifies to (x - 2)² + (y + 5)² = 36. The Pythagorean Theorem makes circle equations!
Two circles intersect at points $A$ and $B$ . The first circle has center $(1, 2)$ and passes through $(4, 6)$ . The second circle has center $(3, 0)$ and the same radius as the first. What is the equation of the line passing through the intersection points $A$ and $B$ ?
Answer: $x - y - 1 = 0$ First, find the radius: $r = \sqrt{(4-1)^2 + (6-2)^2} = \sqrt{9 + 16} = 5$ . Circle 1: $(x-1)^2 + (y-2)^2 = 25$ . Circle 2: $(x-3)^2 + (y-0)^2 = 25$ . The radical axis (line through intersection points) is found by subtracting the equations: $(x-1)^2 + (y-2)^2 - [(x-3)^2 + y^2] = 0$ . Expanding: $x^2 - 2x + 1 + y^2 - 4y + 4 - x^2 + 6x - 9 - y^2 = 0$ . Simplifying: $4x - 4y - 4 = 0$ , or $x - y - 1 = 0$ . Choice A has wrong coefficients. Choice C has wrong constant term. Choice D has wrong sign on constant term.
Complete the square to write the circle equation $x^2+y^2-8x+6y+9=0$ in standard form. Which standard-form equation is correct?
Answer: $(x-4)^2+(y+3)^2=16$ This question tests your understanding of how to derive circle equations using the Pythagorean Theorem (distance formula) or find a circle's center and radius by completing the square. When a circle equation is in general form x² + y² + Dx + Ey + F = 0, we complete the square in both x and y to reveal the center and radius: group x-terms (x² + Dx) and complete the square by adding (D/2)², do the same for y-terms with (E/2)², then rearrange to get (x + D/2)² + (y + E/2)² = (D/2)² + (E/2)² - F. The center is (-D/2, -E/2) and radius is √[(D/2)² + (E/2)² - F]. It's completing the square twice, once for each variable! Starting with x² + y² - 8x + 6y + 9 = 0, move 9 to get x² - 8x + y² + 6y = -9, complete square for x: (x - 4)² - 16, for y: (y + 3)² - 9, so (x - 4)² + (y + 3)² = -9 + 16 + 9 = 16. Choice A correctly completes the square to identify the standard form (x - 4)² + (y + 3)² = 16. A distractor like choice B might forget the signs when completing the square, using +4 for x instead of -4, but gently remember D = -8 so -D/2 = 4 for the center's x-coordinate. Completing the square for circles: (1) Group x-terms together and y-terms together: (x² + Dx) + (y² + Ey) = -F, (2) Complete square in x by adding (D/2)² to both sides, (3) Complete square in y by adding (E/2)² to both sides, (4) Factor: (x + D/2)² + (y + E/2)² = (D/2)² + (E/2)² - F, (5) Read center as (-D/2, -E/2) and radius as √[right side].
Complete the square: $x^2 + y^2 + 4x - 10y + 13 = 0$ . Which standard form equation is correct?
Answer: $(x+2)^2 + (y-5)^2 = 16$ This question tests your understanding of how to derive circle equations using the Pythagorean Theorem (distance formula) or find a circle's center and radius by completing the square. When a circle equation is in general form x² + y² + Dx + Ey + F = 0, we complete the square in both x and y to reveal the center and radius: group x-terms (x² + Dx) and complete the square by adding (D/2)², do the same for y-terms with (E/2)², then rearrange to get (x + D/2)² + (y + E/2)² = (D/2)² + (E/2)² - F. The center is (-D/2, -E/2) and radius is √[(D/2)² + (E/2)² - F]. It's completing the square twice, once for each variable! For x² + y² + 4x - 10y + 13 = 0, it becomes (x + 2)² + (y - 5)² = 16. Choice A correctly completes the square to (x+2)² + (y-5)² = 16. A common issue, as in choice B, is mishandling signs when rewriting the center. Completing the square for circles: (1) Group x-terms together and y-terms together: (x² + Dx) + (y² + Ey) = -F, (2) Complete square in x by adding (D/2)² to both sides, (3) Complete square in y by adding (E/2)² to both sides, (4) Factor: (x + D/2)² + (y + E/2)² = (D/2)² + (E/2)² - F, (5) Read center as (-D/2, -E/2) and radius as √[right side]. Example: x² + y² + 6x - 8y = 0 → (x + 3)² + (y - 4)² = 9 + 16 = 25, so center (-3, 4), radius 5. Super progress—you're a pro at this now!
A cell tower is located at $(3,4)$ on a coordinate grid. Its signal reaches all points within $10$ units of the tower. Write the equation of the circle representing points exactly $10$ units from the tower.
Answer: $(x-3)^2+(y-4)^2=100$ This question tests your understanding of how to derive circle equations using the Pythagorean Theorem (distance formula) or find a circle's center and radius by completing the square. A circle is defined as all points at a fixed distance (radius r) from a center point (h, k): using the distance formula, the distance from any point (x, y) on the circle to the center is √[(x - h)² + (y - k)²] = r. Squaring both sides eliminates the radical and gives the standard form (x - h)² + (y - k)² = r². This equation comes directly from the Pythagorean Theorem applied to the right triangle formed by the horizontal distance (x - h), vertical distance (y - k), and radius r as hypotenuse! For the tower at (3, 4) with reach 10 units, the equation is (x - 3)² + (y - 4)² = 10² = 100. Choice B correctly derives the equation with center (3, 4) and r² = 100. An error like in choice C might flip the signs to (x + 3) and (y + 4), but recall it's (x - h) and (y - k) for positive h and k. Deriving from center and radius: (1) Write the distance from general point (x, y) to center (h, k) using distance formula: √[(x - h)² + (y - k)²], (2) Set equal to radius r, (3) Square both sides to get (x - h)² + (y - k)² = r². That's it! For example, center (2, -5) and radius 6: (x - 2)² + (y - (-5))² = 6², which simplifies to (x - 2)² + (y + 5)² = 36. The Pythagorean Theorem makes circle equations!
The equation $x^2 + y^2 + 4x - 10y + 13 = 0$ can be rewritten in the form $(x - h)^2 + (y - k)^2 = r^2$ . What is the value of $h + k + r$ ?
Answer: 7 Complete the square: $x^2 + 4x + y^2 - 10y = -13$ . For x-terms: $x^2 + 4x + 4 = (x + 2)^2$ . For y-terms: $y^2 - 10y + 25 = (y - 5)^2$ . The equation becomes $(x + 2)^2 + (y - 5)^2 = -13 + 4 + 25 = 16$ . So $h = -2,k = 5$ , and $r = 4$ . Therefore $h + k + r = -2 + 5 + 4 = 7$ . Choice A results from using $r^2$ instead of $r$ . Choice C comes from sign errors with $h$ . Choice D results from calculation errors in completing the square.
All points $(x,y)$ that are $6$ units from the point $(-1,4)$ form a circle. Using the distance formula (from the Pythagorean Theorem), write the equation of this circle in standard form.
Answer: $(x+1)^2+(y-4)^2=36$ This question tests your understanding of how to derive circle equations using the Pythagorean Theorem (distance formula) or find a circle's center and radius by completing the square. A circle is defined as all points at a fixed distance (radius r) from a center point (h, k): using the distance formula, the distance from any point (x, y) on the circle to the center is √[(x - h)² + (y - k)²] = r. Squaring both sides eliminates the radical and gives the standard form (x - h)² + (y - k)² = r². This equation comes directly from the Pythagorean Theorem applied to the right triangle formed by the horizontal distance (x - h), vertical distance (y - k), and radius r as hypotenuse! For points 6 units from (-1, 4), the distance formula gives √[(x + 1)² + (y - 4)²] = 6, squaring to (x + 1)² + (y - 4)² = 36. Choice C correctly derives the equation with the right signs and squared radius of 36. A distractor like choice D might use (y - 4) but flip the x-sign incorrectly, but remember the center's signs: for h = -1, it's (x - (-1))² = (x + 1)². Deriving from center and radius: (1) Write the distance from general point (x, y) to center (h, k) using distance formula: √[(x - h)² + (y - k)²], (2) Set equal to radius r, (3) Square both sides to get (x - h)² + (y - k)² = r². That's it! For example, center (2, -5) and radius 6: (x - 2)² + (y - (-5))² = 6², which simplifies to (x - 2)² + (y + 5)² = 36. The Pythagorean Theorem makes circle equations!
Find the center and radius of the circle given by $x^2+y^2+6x-4y+9=0$ .
Answer: Center $(-3,2)$ , radius $2$ This question tests your understanding of how to derive circle equations using the Pythagorean Theorem (distance formula) or find a circle's center and radius by completing the square. When a circle equation is in general form x² + y² + Dx + Ey + F = 0, we complete the square in both x and y to reveal the center and radius: group x-terms (x² + Dx) and complete the square by adding (D/2)², do the same for y-terms with (E/2)², then rearrange to get (x + D/2)² + (y + E/2)² = (D/2)² + (E/2)² - F. The center is (-D/2, -E/2) and radius is √[(D/2)² + (E/2)² - F]. It's completing the square twice, once for each variable! For x² + y² + 6x - 4y + 9 = 0, move 9: x² + 6x + y² - 4y = -9, complete x: (x + 3)² - 9, y: (y - 2)² - 4, so (x + 3)² + (y - 2)² = -9 + 9 + 4 = 4, center (-3, 2), radius 2. Choice B correctly identifies the center (-3, 2) and radius 2 after completing the square. A mistake like in choice A might not flip the signs for the center, using (3, -2) instead, but remember center is (-D/2, -E/2). Reading center from standard form has a sign trap: in (x - h)² + (y - k)² = r², the center is (h, k), but the signs in the equation are OPPOSITE! From (x - 3)² + (y + 2)² = 16, the center is (3, -2) because (x - 3) has center x = 3, and (y + 2) = (y - (-2)) has center y = -2. Think: what values make each squared term equal zero? Those are your center coordinates. Don't just copy the numbers—flip the signs!
Find the center and radius of the circle $x^2+y^2+6x-4y-12=0$ by completing the square.
Answer: Center $(-3,2)$ , radius $5$ This question tests your understanding of how to derive circle equations using the Pythagorean Theorem (distance formula) or find a circle's center and radius by completing the square. When a circle equation is in general form x² + y² + Dx + Ey + F = 0, we complete the square in both x and y to reveal the center and radius: group x-terms (x² + Dx) and complete the square by adding (D/2)², do the same for y-terms with (E/2)², then rearrange to get (x + D/2)² + (y + E/2)² = (D/2)² + (E/2)² - F. The center is (-D/2, -E/2) and radius is √[(D/2)² + (E/2)² - F]. It's completing the square twice, once for each variable! For x² + y² + 6x - 4y - 12 = 0, group and complete: (x² + 6x) + (y² - 4y) = 12, add 9 and 4 to both sides for (x + 3)² + (y - 2)² = 25, so center (-3, 2) and radius 5. Choice A correctly identifies the center and radius as (-3, 2) and 5. Choice B flips the signs on the center, perhaps by not negating D/2 and E/2 properly. Completing the square for circles: (1) Group x-terms together and y-terms together: (x² + Dx) + (y² + Ey) = -F, (2) Complete square in x by adding (D/2)² to both sides, (3) Complete square in y by adding (E/2)² to both sides, (4) Factor: (x + D/2)² + (y + E/2)² = (D/2)² + (E/2)² - F, (5) Read center as (-D/2, -E/2) and radius as √[right side]. Example: x² + y² + 6x - 8y = 0 → (x + 3)² + (y - 4)² = 25, so center (-3, 4), radius 5.
A circle has the equation $(x - 2)^2 + (y + 1)^2 = 9$ . If this circle is translated so that its center moves to the origin, and then dilated by a factor of $\frac{2}{3}$ centered at the origin, what is the equation of the resulting circle?
Answer: $x^2 + y^2 = 4$ When you encounter circle transformations, you need to apply each transformation step-by-step to see how it affects the equation. The original circle $(x - 2)^2 + (y + 1)^2 = 9$ has center $(2, -1)$ and radius $r = \sqrt{9} = 3$ . First, translating the center to the origin means shifting the circle so its center moves from $(2, -1)$ to $(0, 0)$ . This eliminates the horizontal and vertical shifts in the equation, giving us $x^2 + y^2 = 9$ . The radius remains 3. Next, dilating by a factor of $\frac{2}{3}$ centered at the origin multiplies all distances from the origin by $\frac{2}{3}$ . Since the radius represents distance from the center, the new radius becomes $3 \times \frac{2}{3} = 2$ . Therefore, the final equation is $x^2 + y^2 = 2^2 = 4$ . Looking at the wrong answers: Choice A gives $x^2 + y^2 = \frac{4}{9}$ , which would result from incorrectly squaring the dilation factor $\left(\frac{2}{3}\right)^2$ and multiplying it by the original radius instead of the equation's constant term. Choice B gives $x^2 + y^2 = 6$ , which might come from adding the original radius and new radius ( $3 + 3 \times \frac{2}{3} = 5$ , though this doesn't quite match). Choice C gives $x^2 + y^2 = 9$ , which forgets to apply the dilation entirely. Remember: when dilating a circle, the radius gets multiplied by the dilation factor, so $r^2$ in the equation gets multiplied by the square of that factor.
A circle is tangent to both the x-axis and the line $y = 8$ , and its center has x-coordinate 5. What is the equation of this circle?
Answer: $(x - 5)^2 + (y - 4)^2 = 16$ When you encounter a circle that's tangent to two parallel lines, the key insight is that the circle's center must be equidistant from both lines, and this distance equals the radius. Since the circle is tangent to both the x-axis ( $y = 0$ ) and the line $y = 8$ , its center lies exactly halfway between them. The distance between these parallel lines is 8 units, so the center must be 4 units from each line. This places the center at $y = 4$ . Combined with the given x-coordinate of 5, the center is at $(5, 4)$ . The radius equals the distance from the center to either tangent line. From $(5, 4)$ to the x-axis is 4 units, so $r = 4$ . Since the standard form of a circle is $(x - h)^2 + (y - k)^2 = r^2$ where $(h, k)$ is the center, our equation becomes $(x - 5)^2 + (y - 4)^2 = 16$ . Choice A has the correct radius but places the center at $(5, -4)$ , which would make the circle tangent to $y = 0$ and $y = -8$ , not $y = 8$ . Choice B correctly identifies the center but uses $r^2 = 8$ instead of $r^2 = 16$ —this represents confusing the radius with the distance between the tangent lines. Choice D has the correct radius but incorrectly places the center at $(-5, 4)$ instead of $(5, 4)$ . Remember: when a circle is tangent to two parallel lines, always find the midpoint between them for the center's position perpendicular to those lines.
What are the center and radius of the circle given by $ $(x+5)^2$ + $(y-1)^2$ = 49 $?
Answer: Center $ (-5, 1) $, radius $ 7 $ This question tests your understanding of how to derive circle equations using the Pythagorean Theorem (distance formula) or find a circle's center and radius by completing the square. A circle is defined as all points at a fixed distance (radius $ r $) from a center point ($ h $, $ k $): using the distance formula, the distance from any point ($ x $, $ y $) on the circle to the center is $\sqrt{(x - h)^2 + (y - k)^2} = r$ . Squaring both sides eliminates the radical and gives the standard form $ $(x - h)^2$ + $(y - k)^2$ = $r^2$ $. This equation comes directly from the Pythagorean Theorem applied to the right triangle formed by the horizontal distance ($ x - h $), vertical distance ($ y - k $), and radius $ r $ as hypotenuse! For $ $(x + 5)^2$ + $(y - 1)^2$ = 49 $, rewrite as $ $(x - (-5))^2$ + $(y - 1)^2$ = $7^2$ $, so center $ (-5, 1) $ and radius $ 7 $. Choice B correctly identifies the center with flipped signs and radius as the square root of 49. A mistake like in choice A might ignore the sign trap, using $ (5, -1) $ instead of flipping to $ (-5, 1) $, but remember the center makes each term zero. Reading center from standard form has a sign trap: in $ $(x - h)^2$ + $(y - k)^2$ = $r^2$ $, the center is $ (h, k) $, but the signs in the equation are OPPOSITE! From $ $(x - 3)^2$ + $(y + 2)^2$ = 16 $, the center is $ (3, -2) $ because $ (x - 3) $ has center $ x = 3 $, and $ (y + 2) = (y - (-2)) $ has center $ y = -2 $. Think: what values make each squared term equal zero? Those are your center coordinates. Don't just copy the numbers—flip the signs!

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