Radius (r1) of 1st circle= 19 cmRadius (r2) or 2nd circle = 9\xa0cmLet radius of 3rd circle be rCircumference of 1st circle\xa0{tex}= 2\\pi {r_1} = 2\\pi \\left( {19} \\right) = 38\\pi {/tex}Circumference of 2nd\xa0circle\xa0{tex}= 2\\pi {r_2} = 2\\pi \\left( 9 \\right) = 18\\pi {/tex}Circumference of 3rd circle\xa0{tex} = 2\\pi r{/tex}Given thatCircumference of 3rd circle = circumference of 1st circle + circumference of 2nd circle{tex}2\\pi r = 38\\pi + 18\\pi = 56\\pi {/tex}{tex}r = \\frac{{56\\pi }}{{2\\pi }} = 28{/tex}So, radius of circle which has circumference equal to the sum of the circumference of given two circles is 28 cm.Area of circle\xa0{tex} = \\pi {r^2} = \\left( {\\frac{{22}}{7}} \\right) \\times 28 \\times 28 = 2464c{m^2}{/tex}