Geometry

By tangent-secant theorem: PT² = PA × PB. Given PT = 10.5 and PA = 7: 10.5² = 7 × PB, so 110.25 = 7 × PB, giving PB = 15.75. Chord AB = PB - PA = 15.75 - 7 = 8.75 cm.

In any triangle, the angle formed by the intersection of two angle bisectors is related to the third angle. The formula is: intersection angle = 90° + (third angle)/2. Given the intersection is 112°, we have 112° = 90° + M/2, which gives M/2 = 22°, so M = 44°. This relationship holds because angle bisectors create specific angle relationships in triangles.

When two circles touch externally, the line joining their centers passes through the touch point T. RS is a direct common tangent touching at P and Q. The angle between tangent and radius is 90°, so ∠TPQ and ∠PQT are complementary to the angle formed by the tangent with the line of centers. Using tangent properties: ∠PQT = 90° - ∠TPQ + ∠PTQ relationship. Given ∠TPQ = 42°, and applying circle theorems about tangents from external points, we get ∠PQT = 48°.

For points on a circle, the angle at the center is twice the angle at the circumference subtended by the same chord. Central angle AOB = 110° and AOC = 130°, so angle BOC = 360° - 110° - 130° = 120° (assuming points are arranged around the circle). Since angle BAC subtends chord BC at the circumference, it equals half of angle BOC: 120° / 2 = 60°.

In a circle with diameter PQ, when chords PR and QS meet at T externally, angle PTQ equals half the sum of arcs intercepted by the angle. Given ∠ROS = 42°, the central angle subtended by chord RS is 42°. The angle PTQ is half the difference of 180° and 42°, which equals 69°. Alternatively, using cyclic quadrilateral properties: ∠PTQ = 90° - (∠ROS/2) = 90° - 21° = 69°. Option D is correct.

For a tangent to a circle from an external point, the tangent is perpendicular to the radius at the point of contact. Using the tangent-secant theorem: TA² = OT² - OA². Given OT = 26 cm and OA = radius = 10 cm (diameter = 20 cm), we get TA² = 26² - 10² = 676 - 100 = 576. Therefore, TA = 24 cm. Option C is correct.

For a square inscribed in a circle, the diagonal of the square equals the diameter of the circle. Radius = 5 cm, so diameter = 10 cm. If side = a, then diagonal = a√2 = 10, giving a = 10/√2 = 5√2 cm. Area of square = a² = (5√2)² = 25 × 2 = 50 cm².

For a circle of radius 10 cm (half of diameter 20 cm), the perpendicular distance from center to chord AB is 6 cm. By chord length formula: AB = 2√(r² - d²) = 2√(100 - 36) = 2√64 = 16 cm. Since BC is diameter and AB ∥ CD, chords AB and CD are symmetric about the center. CD is 16 cm - 6 cm = 10 cm from center on the opposite side, giving the same length of 16 cm.

By the Power of a Point theorem for intersecting chords: PA × PB = PC × PD. Here PA = AB + BP = 6.3 + 4.5 = 10.8 cm, PB = 4.5 cm. Therefore 10.8 × 4.5 = 3.6 × PD, giving PD = (10.8 × 4.5)/3.6 = 48.6/3.6 = 13.5 cm. But wait - let me recalculate: 10.8 × 4.5 = 48.6, and 48.6/3.6 = 13.5. Hmm, let me verify: PA × PB = PC × PD means 10.8 × 4.5 = (CD + PD) × PD = (3.6 + PD) × PD. This gives PD² + 3.6PD - 48.6 = 0. Solving: PD = [-3.6 + √(12.96 + 194.4)]/2 = [-3.6 + √207.36]/2 = [-3.6 + 14.4]/2 = 5.4 cm.

For a square inscribed in a circle, the diagonal of the square equals the diameter of the circle. Diameter = 24 cm, so diagonal = 24. If side = s, then s√2 = 24, so s = 24/√2 = 12√2. Area = s^2 = (12√2)^2 = 288 cm^2.

For two circles touching externally with radii R and r, direct common tangent length = 2√(Rr). Here R = 36, r = 16, so length = 2√(36 × 16) = 2√576 = 2 × 24 = 48 cm.

∠CBD = 180° - ∠ABC and ∠BCE = 180° - ∠ACB (exterior angles). Since ∠A + ∠ABC + ∠ACB = 180° and ∠A = 66°, we have ∠ABC + ∠ACB = 114°. In quadrilateral BOC formed by bisectors: ∠BOC = 180° - (∠CBD/2 + ∠BCE/2) = 180° - (180° - ∠A)/2 = 180° - 114°/2 = 180° - 57° = 57°.

10² + 24² = 100 + 576 = 676 = 26², confirming right triangle. In a right triangle, the median to the hypotenuse equals half the hypotenuse: 26/2 = 13 cm.

For chords in concentric circles, if a line cuts both circles, the perpendicular distance from the center to the chord is the same for both circles. Let this distance be d. For the smaller circle: (4/2)² + d² = r², so r² = 4 + d². For the larger circle: (16/2)² + d² = R², so R² = 64 + d². Difference in squares of radii = R² - r² = (64+d²) - (4+d²) = 60.

When two chords intersect at point P inside a circle, the angle between them (∠APC = 40°) is half the sum of the measures of the arcs intercepted by the angle and its vertical angle. The central angles ∠AOC and ∠BOD subtend these arcs. The relationship is: ∠AOC + ∠BOD = 2 × ∠APC = 2 × 40° = 80°.