Let the radius of the bigger circle be R cm. and radii of two smaller circles are r₁and r₂, then according to question.
2πR = 2πr₁ + 2πr₂
⇒    2πR = 2π (19) + 2π (9)
⇒    2πR = 2π (19 + 9)
⇒    R = 28
Hence, radius of the circle be 28 cm.

Let the radius of the required circle be R and radii of two given circles be r₁ and r₂, then according to question.
πR² = πr₁² + πr₂²
⇒    πR² = (r₁² + r₂²)
⇒    R² = 64 + 36
⇒    R² = 100
⇒    R = 10 cm.

We have,
r = Radius of the region representing Gold score = 10.5 cm
∴ r₁ = Radius of the region representing Gold and Red scoring areas = (10.5 + 10.5) cm = 21 cm = 2r cm
r₂ = Radius of the region representing Gold, Red and Blue scoring areas = (21 + 10.5) cm = 31.5 cm = 3r cm
r₃ = Radius of the region representing Gold, Red, Blue and Black scoring areas = (31.5 + 10.5) cm = 42 cm = 4r cm
r₄ = Radius of the region representing Gold, Red, Blue, Black and white scoring areas = (42 + 10.5) cm = 52.5 cm = 5r cm
Now, A, = Area of the region representing Gold scoring area

= 22 × 1.5 × 10.5 = 346.5 cm²
A₂ = Area of the region representing Red scorring area
= π (2r)² - πr² = 3πr² = 3A₁
= 3 × 346.5 cm² = 1039.5 cm²
A₃ = Area of the region representing Blue scoring area
= π(3r)² - π(2r)² = 9πr² - 4πr²
= 5πr² = 5A₁ = 5 × 346.5 cm²
= 1732.5 cm²
A₄ = Area of the region representing black scoring area
= π(4πr)² - π(3r)² = 7πr² = 7A₁
=7 × 346.5 cm² = 2425.5 cm²
A₁ = Area of the region representing white scoring area
= π(5r)² - π(4r)² - 9πr² = 9 A₁
= 9 × 346.5 cm² = 3118.5 cm²