A right conical cap just covers two spheres placed one above the other on a table such that it touches both the spheres. Let r be the radius of the smaller sphere and R be the radius of the bigger sphere. Let 2θ be the vertical angle of the cone. What is the radius of the base of the cone?
- A2r²tan θ/(R − r)
- B2R²tan θ/(R − r)
- C2(r² + R²)tan θ/(R − r)
- D(r² + R²)tan θ/(R − r)
Explanation
In an axial cross-section the cone is an isosceles triangle of semi-vertical angle θ and each sphere is a circle of radius r or R tangent to both sides. If A is the apex, a circle of radius ρ tangent to both sides has its centre at distance ρ/sin θ from A. The spheres touch each other, so the gap between their centres is R + r, giving (R − r)/sin θ = R + r, i.e. sin θ = (R − r)/(R + r). The cap just covers both spheres with its base on the table, so the cone height equals the apex height above the table: the bigger centre is R above the table, hence H = R/sin θ + R = R(1 + sin θ)/sin θ. The base radius is H·tan θ = R(1 + sin θ)/cos θ, and since 1 + sin θ = 2R/(R + r), this equals 2R²·tan θ/(R − r), option (B).