Geometry

For two circles of radii r1 and r2 intersecting orthogonally, the distance between centers d = sqrt(r1^2 + r2^2) = sqrt(3^2 + 4^2) = 5. The length of the common chord is (2 * r1 * r2) / d = (2 * 3 * 4) / 5 = 24/5.

The circle equation is (x-8)^2 + (y-6)^2 = 36. i: Polar of (-5, 1) is (x+5)(8) + (y-1)(6) - 8(x-5) - 6(y+1) + 64 = 0, which simplifies to 13x + 5y = 98. ii: Tangent at (8, 0) is (x-8)(0) + (y-6)(-6) = 36, which is y=0. iii: Normal at (2, 6) is y=6. iv: Diameter through (8, 12) is x=8.

Area of isosceles right triangle = (1/2) * side^2 = 200, so side^2 = 400, side = 20. Hypotenuse = sqrt(side^2 + side^2) = sqrt(400 + 400) = sqrt(800) = 20 * sqrt(2).

The radical center of three circles is the point from which the lengths of tangents to all three circles are equal. A circle centered at the radical center with this tangent length as the radius is the radical circle, which is orthogonal to all three circles.

The locus of points from which tangents to two circles are equal is the radical axis. For three circles, the intersection of the radical axes of the pairs of circles is the radical center, which is a single point.

Vertices are (6,6), (0,6), (6,0). This is a right triangle. Circumcentre is the midpoint of the hypotenuse: ((6+0)/2, (6+6)/2) = (3,6) is wrong, it's (3,3). Centroid is ((6+0+6)/3, (6+6+0)/3) = (4,4). Distance = sqrt((4-3)^2 + (4-3)^2) = sqrt(1+1) = sqrt(2).

The circle is (x-2)^2 + (y-1)^2 = 11 + 4 + 1 = 16. Radius r = 4. Distance from (4,5) to center (2,1) is sqrt((4-2)^2 + (5-1)^2) = sqrt(4+16) = sqrt(20). Length of tangent L = sqrt(d^2 - r^2) = sqrt(20 - 16) = 2. The area of the quadrilateral formed by two tangents and two radii is 2 * (1/2 * L * r) = L * r = 2 * 4 = 8.

By the Law of Sines, a = 2R sin(A). Differentiating with respect to time, da/dt = 2R cos(A) * dA/dt. Given da/dt = R * dA/dt, we have 2R cos(A) = R, so cos(A) = 1/2. Thus, A = 60 degrees or pi/3.

For a chord QR parallel to the tangent at P, the triangle PQR has its maximum area when the height from P to the chord QR is maximized. This occurs when the triangle is equilateral, with height 3r/2. The area is (1/2) * base * height, which equals (3 * sqrt(3) / 4) * r^2.

For a triangle inscribed in a semicircle with the diameter as one side, the height is maximized when the vertex is at the radius perpendicular to the diameter. Max height = radius = a. Base = diameter = 2a. Area = (1/2) * base * height = (1/2) * 2a * a = a^2.

A rectangle inscribed in a circle has maximum area when it is a square. Diagonal of square = diameter of circle = 10 cm. Side s = 10 / sqrt(2). Area = s^2 = 100 / 2 = 50 sq cm.

For a circle of radius R, the triangle of maximum area inscribed is an equilateral triangle. The side length of this triangle is R*sqrt(3), and its in-radius r is given by R/2. Given R=46, the in-radius is 46/2 = 23.

For a rectangle inscribed in a semicircle of radius R, the area is maximized when the width is R * sqrt(2) and height is R / sqrt(2). With R = 8, width = 8 * sqrt(2) and height = 8 / sqrt(2) = 4 * sqrt(2).