Question

Casey’s theorem proof step by step

Answer 1

Two circles Γ1(r1) and Γ2(r2) are internally/externally tangent to a circle Γ(R) through
A, B, respetively. The length δ12 of the common external tangent of Γ1, Γ2 is given by:
δ12 = AB/R √{(R ± r1)(R ± r2)}

Proof. Without loss of generality assume that r1 ≥ r2 and we suppose that Γ1 and Γ2 are internally
tangent to Γ. The remaining case will be treated analogously. A common external tangent between Γ1
and Γ2 touches Γ1, Γ2 at A1, B1 and A2 is the orthogonal projection of O2 onto O1A1.
By Pythagorean theorem for ∆O1O2A2, we obtain

δ12^2 = (A1B1)^2 = (O1O2)^2 − (r1 − r2)^2

Let < O1OO2 = λ. By cosine law for ∆OO1O2, we get
(O1O2)^2 = (R − r1)^2 + (R − r2)^2 − 2(R − r1)(R − r2) cos λ
By cosine law for the isosceles triangle 4OAB, we get
AB^2 = 2R^2(1 − cos λ)

Eliminating cos λ and O1O2 from the three previous expressions yields
δ12^2 = (R − r1)^2 + (R − r2)^2 − (r1 − r2)^2 − 2(R − r1)(R − r2)( 1 −AB^2 /2R2)

Subsequent simplifications give
δ12 =AB/R√{(R − r1)(R − r2) }
Analogously, if Γ1, Γ2 are externally tangent to Γ, then we will get
δ12 =AB/R√{(R + r1)(R + r2)}
If Γ1 is externally tangent to Γ and Γ2 is internally tangent to Γ, then a similar reasoning gives that
the length of the common internal tangent between Γ1 and Γ2 is given by
δ12 =AB/R√{(R + r1)(R − r2) }