Radius of first circle (r₁) = 8 cm,
Radius of second circle (r₂) = 6 cm
Let the radius of the third circle = r
Area of the first circle = πr₁² = π(64) = 64π
Area of the second circle = πr₂² = π(36) = 36π
Area of the third circle = πr²
According to question,
Area of the third circle = Area of the first circle + Area of the second circle
⇒ πr² = 64π + 36π ⇒ πr² = 100π ⇒ r = √100 = 10 cm
Hence, the radius of the circle, having area equal to the sum of the areas of the two circles, is 10 cm.

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