Geometry

The perpendicular from the center to a chord bisects the chord. This forms a right triangle with the radius as the hypotenuse, half the chord (7 cm) as one leg, and the distance from the center (24 cm) as the other leg. Radius = sqrt(7^2 + 24^2) = sqrt(49 + 576) = sqrt(625) = 25.

Let r be the radius. The distance from the center to chord AB is sqrt(r^2 - 40^2) and to chord CD is sqrt(r^2 - 9^2). Since they are on the same side, sqrt(r^2 - 81) - sqrt(r^2 - 1600) = 31. Solving this equation for r yields r = 41.

If chord length equals radius, the triangle formed by the chord and the center is equilateral. The angle at the center is 60 degrees. The angle subtended by the chord on the major arc is half the angle at the center, which is 30 degrees.

The common chord of (x-a)^2 + y^2 = a^2 and x^2 + (y+b)^2 = b^2 is found by subtracting the equations: x^2 - 2ax + a^2 + y^2 = a^2 and x^2 + y^2 + 2by + b^2 = b^2. x^2 + y^2 - 2ax = 0 and x^2 + y^2 + 2by = 0. Subtracting gives 2ax + 2by = 0, or ax + by = 0. The circle with this chord as diameter is (x-x1)(x-x2) + (y-y1)(y-y2) = 0. The derived equation matches option C.

The first circle has center (0,0) and radius 2. The second circle has center (4,0) and radius 2. The distance between centers is 4, which equals the sum of the radii (2+2), meaning the circles touch externally; therefore, they have 3 common tangents.

In a right triangle with legs a and b, the medians m1 and m2 from the acute angles satisfy 4(m1^2 + m2^2) = 5c^2, where c is the hypotenuse. Substituting m1=6 and m2=8 gives 4(36 + 64) = 5c^2, so 400 = 5c^2, c^2 = 80, and c = sqrt(80) = 4*sqrt(5).

Area of a triangle with two sides r and included angle theta is (1/2) * r^2 * sin(theta). Here r=14, theta=90 degrees. Area = (1/2) * 14 * 14 * sin(90) = 0.5 * 196 * 1 = 98.

The triangle formed by the chord and the center is an isosceles triangle with sides 10, 10, and 12. By drawing an altitude to the chord, we split it into two right triangles with hypotenuse 10 and base 6, making the height 8 via the Pythagorean theorem. The area is 1/2 * base * height = 1/2 * 12 * 8 = 48 sq. cm.

The triangle is a right-angled triangle (9^2 + 12^2 = 81 + 144 = 225 = 15^2). The inradius of a right triangle is (a+b-c)/2 = (9+12-15)/2 = 6/2 = 3.

The chord with midpoint (h,k) has equation xh + yk = h^2 + k^2. Since it passes through (x1, y1), x1h + y1k = h^2 + k^2. Replacing (h,k) with (x,y) gives x^2 + y^2 - xx1 - yy1 = 0.

A right triangle is formed by the radius of the smaller circle (5 cm), half the chord, and the radius of the larger circle (13 cm). Using Pythagoras, half-chord = sqrt(13^2 - 5^2) = sqrt(169 - 25) = sqrt(144) = 12. The full chord length is 12 * 2 = 24 cm.

The radii are 9 and 12, and the distance between centers is 15. Since 9^2 + 12^2 = 81 + 144 = 225 = 15^2, the triangle formed by the centers and the intersection point is a right triangle. The common chord is twice the altitude to the hypotenuse. Altitude = (9 * 12) / 15 = 7.2. Chord = 2 * 7.2 = 14.4.

For two circles touching at X with common external tangents, the distance from the intersection of tangents P to the point of contact is given by properties of tangents. Using similar triangles or the property that the tangent length squared equals the product of distances, the radius of the smaller circle is 3.75 cm.

For two circles touching at X with a common tangent YZ, the tangent at X bisects YZ. Thus, YZ = 2 * XA. Given XA = 16, YZ = 32.