Given the lengths of the radii of two intersecting Circles, and the distance between their centres: find the area of the Tetragon formed by the tangents at the points of intersection.
If ‘2M’ = area of Tetragon whose vertices are the Centres and the Points of intersection; and if its sides be ‘a’, ‘b’, and its diagonal, joining the Centres, ‘c’: required area
32M³ |
(b² + c² − a²) · (c² + a² − b²) |
Let A, B, be the centres of the Circles; C, D their points of intersection; and CFDE the Tetragon whose area is required.
Let the sides of the Triangle ABC be a, b, c; and its ∠s α, β, γ.
Then CE = b · tan α, and CF = a · tan β.
Also ∠FCE = ∠ACE + ∠FCB - γ = π − γ;
∴ sin FCE = sin γ.
Hence area of Triangle FCE = ½ · ab · tan α · tan β · sin γ;
∴ area of Tetragon =
ab sin α sin β sin γ |
cos α cos β |
Now, writing ‘M’ for area of Triangle ABC, we have
2M |
bc |
2M |
ca |
2M |
ab |
∴ area of Tetragon = ab ·
8M³ |
a²b²c² |
4bc · ca |
(b² + c² − a²) · (c² + a² − b²) |
32M³ |
(b² + c² − a²) · (c² + a² − b²) |
Q.E.F.