Consider the figure shown below where O is centre of circle, join BC which passes through O, let the side of square be ‘a’ and radius of circle be ‘r’.

Now we know OB and OC are radius of circle

So, OB = OC = r

Consider ∆BDC right angled at D

And we know BC = OC + OB

BC = 2r and BD = DC = a (put these values in eqn1)

⇒ (2r)² = a² + a²

⇒ 4r² = 2a²

Area of inscribed square = side × side

Areaa of inscribed square = a × a

Area of inscribed square = a²→ eqn3

Area of circumscribing circle = πR² where R = radius of circle

⇒ Area of circumscribing circle = πr²→ eqn4

Put the values from equation 3 & 4 in above equation

(from eqn 2)

So, Ratio is π : 2

The ratio is π:2