Geometry

For two circles with radii r₁=7 cm and r₂=2 cm, distance between centers d=13 cm. Length of direct common tangent = √(d² - (r₁ - r₂)²) = √(13² - (7-2)²) = √(169 - 25) = √144 = 12 cm.

For two circles with radii r₁=13 cm and r₂=7 cm, distance between centers d=25 cm. Length of transverse common tangent = √(d² - (r₁ + r₂)²) = √(25² - (13+7)²) = √(625 - 400) = √225 = 15 cm.

AOC is diameter, so ∠ABC = 90° (angle in semicircle). Given ∠AOB = 130° (assuming 130°, not 1300 which is clearly a typo). ∠BOC = 180° - 130° = 50°. BD is chord meeting OB at... We need ∠BDC. In cyclic quadrilateral or using circle properties: ∠BDC = 180° - ∠BOC = 180° - 50° = 130°, but this isn't an option. Using inscribed angle theorem: ∠BDC = ∠BOC/2 = 25°.

The arc length formula is (θ/360) × 2πr, where θ is the central angle in degrees. Substituting θ=72° and r=35 cm: (72/360) × 2π × 35 = (1/5) × 70π = 14π ≈ 43.98 cm, which rounds to 44 cm. Option A (36 cm), B (33 cm), and D (46 cm) are incorrect calculations.

When two circles touch externally at point P and AB is a common tangent, the radii to the points of tangency are perpendicular to the tangent. The angle between the two radii at P is supplementary to angle APB. For externally tangent circles, this angle is always 90°, making ∠APB a right angle.

For a square inscribed in a circle of radius r: diagonal = 2r, so side = 2r/√2 and area = 2r². For a square inscribed in a semicircle of radius r: if the diameter is the base, side = r/√2 and area = r²/2. Ratio of areas = 2r²:(r²/2) = 4:1. The question likely expects the square on diameter configuration.

Arc length = radius × angle (in radians). For equal arcs: r₁ × 45° = r₂ × 75°, so r₁/r₂ = 75/45 = 5/3. The ratio of radii is inverse to the ratio of angles when arc lengths are equal.

Using tangent-secant theorem: PT² = PB × PC. PT = √(10² - 6²) = 8. So 64 = 5 × PC, giving PC = 12.8. Chord BC = PC - PB = 12.8 - 5 = 7.8 cm. Option C is correct.

PA and PB are equal tangents from external point P to the circle. The radius OA is perpendicular to tangent PA at the point of contact. OP bisects the angle between the two equal tangents, so angle APO = 40 degrees. In right triangle APO, angle POA = 180 - 90 - 40 = 50 degrees.

For two circles of radii a and b touching externally, with a third circle of radius c touching both circles and their common external tangent, the relationship is derived from the formula for a Soddy circle: 1/√c = 1/√a + 1/√b. Rearranging gives √c = √ab/(√a + √b). Option D (√bc + √ac = √ab) is algebraically equivalent to this relationship and is the correct answer.

For two circles with radii 9 cm and 16 cm, and center distance 25 cm, the length of the common external tangent is √(d² - (r₁ - r₂)²) = √(25² - (16-9)²) = √(625 - 49) = √576 = 24 cm. 12 and 18 are incorrect calculations, while 21 doesn't match the formula result.

Since PA and PB are tangents to the circle, OA ⟂ PA and OB ⟂ PB. In quadrilateral PAOB, ∠OAP = ∠OBP = 90°. Sum of angles = 360°, so ∠AOB + ∠APB + 90° + 90° = 360°, giving ∠AOB + ∠APB = 180°. Given the ratio 5:1, let ∠AOB = 5x and ∠APB = x. Then 6x = 180°, so x = 30°.

For a triangle with centroid G, AG:GD = 2:1 in median AD. If AG = BC, then using Apollonius theorem and centroid properties, we can derive that ∠BGC = 90°. This occurs when the triangle has a specific relationship between its centroid and side lengths.

For chord length 16 cm at 15 cm from center: r² = 8² + 15² = 64 + 225 = 289, so r = 17 cm. For chord at 8 cm from center: chord/2 = √(17² - 8²) = √(289 - 64) = √225 = 15 cm. Therefore chord = 30 cm.

Larger radius = 5 cm, smaller radius = 3 cm. For a chord of the larger circle tangent to the smaller one, the perpendicular distance from center to chord equals the smaller radius (3 cm). By the Pythagorean theorem, half-chord = √(5² - 3²) = √16 = 4 cm. Full chord length = 2 × 4 = 8 cm.