If d₁, d₂ (d₂<d₁) are the diameters of two concentric circles and chord of one circle of length C is tangent to other circle, then prove that d₂² = C² + d₁².

Let the two concentric circles have the centre O and let AB be the chord of an outer circle whose length is D and which is also tangent to the inner circle at point D as shown

The diameters are given as d₁ and d₂ hence the radius will be and

In ΔOAB

⇒ OA = OB …radius of the outer circle

Hence ΔOAB is an isosceles triangle

As radius is perpendicular to tangent OC is perpendicular to AB

OC is altitude from the apex, and in an isosceles triangle, the altitude is also the median

Hence

Consider ΔODB

⇒ ∠ODB = 90° …radius perpendicular to tangent

Using Pythagoras theorem

⇒ OD² + BD² = OB²

Multiply the whole by 2²

⇒ d₂² + C² = d₁²

Hence proved