Daily Olympiad: Math - Circles [20260607]

Solution

Correct: A

The standard equation of a circle with center (h, k) and radius r is (x - h)^2 + (y - k)^2 = r^2. Given the center (h, k) = (3, -2), the equation starts as (x - 3)^2 + (y - (-2))^2 = r^2, which simplifies to (x - 3)^2 + (y + 2)^2 = r^2. To find r^2, we use the fact that the circle passes through the point (7, 1). We can substitute these coordinates into the equation: (7 - 3)^2 + (1 + 2)^2 = r^2 (4)^2 + (3)^2 = r^2 16 + 9 = r^2 25 = r^2 So, the equation of the circle is (x - 3)^2 + (y + 2)^2 = 25.

Solution

Correct: A

To find the center and radius, we need to rewrite the equation in standard form (x - h)^2 + (y - k)^2 = r^2 by completing the square. Group the x-terms and y-terms: (x^2 - 6x) + (y^2 + 4y) = 12 Complete the square for x: (x^2 - 6x + (-6/2)^2) = (x^2 - 6x + 9) Complete the square for y: (y^2 + 4y + (4/2)^2) = (y^2 + 4y + 4) Add the constants added to both sides of the equation: (x^2 - 6x + 9) + (y^2 + 4y + 4) = 12 + 9 + 4 (x - 3)^2 + (y + 2)^2 = 25 From the standard form, the center (h, k) is (3, -2) and r^2 = 25, so the radius r = sqrt(25) = 5.

Solution

Correct: B

First, find the center of the circle, which is the midpoint of the diameter. Midpoint M = ((x1 + x2)/2, (y1 + y2)/2) M = ((-1 + 5)/2, (5 + (-3))/2) = (4/2, 2/2) = (2, 1). Next, find the radius. The radius is the distance from the center to one of the endpoints. Let's use (5, -3). Radius r = sqrt((5 - 2)^2 + (-3 - 1)^2) r = sqrt((3)^2 + (-4)^2) r = sqrt(9 + 16) r = sqrt(25) r = 5 Finally, calculate the area of the circle, A = πr^2. A = π(5)^2 = 25π.

Solution

Correct: A

Since the point P(4, k) lies on the circle, its coordinates must satisfy the circle's equation. Substitute x = 4 and y = k into the equation: (4 - 1)^2 + (k + 2)^2 = 34 (3)^2 + (k + 2)^2 = 34 9 + (k + 2)^2 = 34 (k + 2)^2 = 34 - 9 (k + 2)^2 = 25 Take the square root of both sides: k + 2 = ±sqrt(25) k + 2 = ±5 Two possible cases: 1) k + 2 = 5 => k = 5 - 2 => k = 3 2) k + 2 = -5 => k = -5 - 2 => k = -7 So, the possible values of k are -7 and 3.

Solution

Correct: C

The formula for the circumference of a circle is C = 2πr, where r is the radius. Given C = 30π feet, we can find the radius: 30π = 2πr Divide both sides by 2π: r = 30π / 2π r = 15 feet The formula for the area of a circle is A = πr^2. Substitute r = 15 into the area formula: A = π(15)^2 A = 225π square feet.

Solution

Correct: C

The formula for the length of an arc (L) is L = (θ/360°) * 2πr, where θ is the central angle in degrees and r is the radius. Given θ = 120 degrees and r = 9 units. L = (120/360) * 2π(9) L = (1/3) * 18π L = 6π units.

Solution

Correct: C

The formula for the area of a sector (A_sector) is A_sector = (θ/360°) * πr^2, where θ is the central angle in degrees and r is the radius. Given A_sector = 18π and θ = 45 degrees. 18π = (45/360) * πr^2 Simplify the fraction: 45/360 = 1/8 18π = (1/8) * πr^2 Divide both sides by π: 18 = (1/8) * r^2 Multiply both sides by 8: r^2 = 18 * 8 r^2 = 144 Take the square root: r = sqrt(144) r = 12 units. (Radius must be positive).

Solution

Correct: B

The measure of an inscribed angle is half the measure of its intercepted arc. Given the intercepted arc measure is 80 degrees. Inscribed angle = (1/2) * 80 degrees = 40 degrees.

Solution

Correct: D

When a radius (or part of it) is drawn perpendicular to a chord, it bisects the chord. This forms a right-angled triangle where: - The hypotenuse is the radius (r = 13 cm). - One leg is half the length of the chord (10 cm / 2 = 5 cm). - The other leg is the perpendicular distance from the center to the chord (let's call it d). Using the Pythagorean theorem: a^2 + b^2 = c^2 5^2 + d^2 = 13^2 25 + d^2 = 169 d^2 = 169 - 25 d^2 = 144 d = sqrt(144) d = 12 cm.

Solution

Correct: C

When two circles are tangent externally, the distance between their centers is equal to the sum of their radii. Let r1 be the radius of the first circle and r2 be the radius of the second circle. r1 = 5 units r2 = 3 units Distance between centers = r1 + r2 = 5 + 3 = 8 units.

Solution

Correct: C

The equation of the circle centered at the origin with radius 6 is x^2 + y^2 = 6^2, which is x^2 + y^2 = 36. Substitute the equation of the line y = -x + 6 into the circle's equation: x^2 + (-x + 6)^2 = 36 x^2 + (x^2 - 12x + 36) = 36 2x^2 - 12x + 36 = 36 2x^2 - 12x = 0 Factor out 2x: 2x(x - 6) = 0 This gives two possible values for x: 2x = 0 => x = 0 x - 6 = 0 => x = 6 For each x-value, find the corresponding y-value using y = -x + 6: If x = 0, y = -0 + 6 = 6. So, (0, 6) is an intersection point. If x = 6, y = -6 + 6 = 0. So, (6, 0) is an intersection point. Since there are two distinct (x, y) solutions, there are 2 points of intersection.

Solution

Correct: B

The equation of the circle is (x - 2)^2 + (y + 1)^2 = 9. This means the center of the circle is (2, -1) and the radius squared is r^2 = 9, so r = 3. To determine the relationship of the point (5, 0) to the circle, we calculate the distance from the center (2, -1) to the point (5, 0). Distance d = sqrt((x2 - x1)^2 + (y2 - y1)^2) d = sqrt((5 - 2)^2 + (0 - (-1))^2) d = sqrt((3)^2 + (1)^2) d = sqrt(9 + 1) d = sqrt(10) Now, compare this distance to the radius (r = 3). Since sqrt(10) is approximately 3.16, and 3.16 > 3, the distance from the center to the point is greater than the radius. Therefore, the point (5, 0) is outside the circle.

Solution

Correct: C

The center of the circle is (h, k) = (1, 3). If the circle is tangent to the x-axis, it means the distance from the center to the x-axis is equal to the radius. The x-axis is the line y = 0. The distance from a point (h, k) to the line y = 0 is |k|. In this case, r = |3| = 3. Now, substitute the center (1, 3) and radius r = 3 into the standard circle equation (x - h)^2 + (y - k)^2 = r^2: (x - 1)^2 + (y - 3)^2 = 3^2 (x - 1)^2 + (y - 3)^2 = 9.

Solution

Correct: A

The original circle has its center at (h, k) = (-4, 5) and radius r = sqrt(16) = 4. When a circle is translated: - 3 units to the right means adding 3 to the x-coordinate of the center: New x-coordinate = -4 + 3 = -1. - 2 units down means subtracting 2 from the y-coordinate of the center: New y-coordinate = 5 - 2 = 3. The new center is (-1, 3). The radius remains unchanged during a translation. So, the new equation of the circle is (x - (-1))^2 + (y - 3)^2 = 16, which simplifies to (x + 1)^2 + (y - 3)^2 = 16.

Solution

Correct: C

The area of the circle is given as A_circle = 144π. The formula for the area of a sector (A_sector) is A_sector = (θ/360°) * A_circle, where θ is the central angle in degrees. Given θ = 75 degrees and A_circle = 144π. A_sector = (75/360) * 144π Simplify the fraction 75/360. Both are divisible by 15: 75/15 = 5, 360/15 = 24. So, 75/360 = 5/24. A_sector = (5/24) * 144π Since 144 / 24 = 6: A_sector = 5 * 6π A_sector = 30π square units.

Solution

Correct: B

The standard equation of a circle with center (h, k) is (x - h)^2 + (y - k)^2 = r^2. Given the center (h, k) = (3, 4), the equation starts as (x - 3)^2 + (y - 4)^2 = r^2. Since the circle passes through the origin (0, 0), we can use these coordinates to find r^2. Substitute x = 0 and y = 0 into the equation: (0 - 3)^2 + (0 - 4)^2 = r^2 (-3)^2 + (-4)^2 = r^2 9 + 16 = r^2 25 = r^2 So, the equation of the circle is (x - 3)^2 + (y - 4)^2 = 25.

Solution

Correct: C

When a regular hexagon is inscribed in a circle, the side length of the hexagon is equal to the radius of the circle. This is because a regular hexagon can be divided into six equilateral triangles, each with vertices at the center of the circle and two adjacent vertices of the hexagon. The sides of these equilateral triangles are all equal to the radius. Given the radius (r) = 8 units. The side length (s) of the regular hexagon = r = 8 units. A hexagon has 6 equal sides. The perimeter of the hexagon = 6 * s = 6 * 8 = 48 units.

Solution

Correct: A

First, find the center of the circle, which is the midpoint of the diameter. Center (h, k) = ((1 + 7)/2, (6 + 6)/2) = (8/2, 12/2) = (4, 6). Next, find the radius. The radius is the distance from the center to one of the endpoints. Let's use (1, 6). Radius r = sqrt((1 - 4)^2 + (6 - 6)^2) r = sqrt((-3)^2 + (0)^2) r = sqrt(9) = 3. The equation of the circle is (x - 4)^2 + (y - 6)^2 = 3^2, or (x - 4)^2 + (y - 6)^2 = 9. Now, check each choice: A) For (4, 9): (4 - 4)^2 + (9 - 6)^2 = 0^2 + 3^2 = 0 + 9 = 9. This point is on the circle. B) For (1, 3): (1 - 4)^2 + (3 - 6)^2 = (-3)^2 + (-3)^2 = 9 + 9 = 18 ≠ 9. C) For (7, 3): (7 - 4)^2 + (3 - 6)^2 = (3)^2 + (-3)^2 = 9 + 9 = 18 ≠ 9. D) For (4, 6): This is the center of the circle, not a point on the circle (unless the radius is 0, which it isn't). (4-4)^2 + (6-6)^2 = 0+0 = 0 ≠ 9. Therefore, (4, 9) is on the circle.

Solution

Correct: B

First, find the radius of the circle from its circumference. C = 2πr 10π = 2πr r = 5 When a square is inscribed in a circle, the diagonal of the square is equal to the diameter of the circle. Diameter d = 2r = 2 * 5 = 10. Let 's' be the side length of the square. By the Pythagorean theorem, the diagonal of a square is s*sqrt(2). So, s*sqrt(2) = 10 s = 10 / sqrt(2) s = (10*sqrt(2)) / 2 s = 5*sqrt(2) The area of the square is A = s^2. A = (5*sqrt(2))^2 A = 5^2 * (sqrt(2))^2 A = 25 * 2 A = 50 square units.

Solution

Correct: A

The area of a circle is given by the formula A = πr^2. We are given that the area of the circle is 49π. Set the formula equal to the given area: πr^2 = 49π Divide both sides by π: r^2 = 49 Take the square root of both sides to find r. Since r represents a radius, it must be positive: r = sqrt(49) r = 7.