Question

answers plssss.......!!!

Answer 1

Let r1 and r2 are the radius of two circle.

Sum of their areas are


$\pi \ r1 {}^{2} + \pi \ r2 {}^{2} = 130\pi \: \: \: \: \: \: \: \: eq1$
Distance between there centres are

$r1 - r2 = 8 \: \: \: \: \: \: \: eq2$

$r1 = 8 + r2$

$now \: put \: value \: of \: r1 \: in \: eq1$

$\pi \ (8 + r2) {}^{2} + \pi \ r 2 {}^{2} = 130\pi$

$cancel \: \pi \: on \: both \: side$

$64 + 16r2 \: + r2 {}^{2} + r2 {}^{2} = 130$

$2r2 {}^{2} + 16r2 - 66 = 0$

Take 2 common

$r2 {}^{2} + 8r2 - 33 = 0$

$r2 {}^{2} + 11r2 - 3r2 - 33 = 0$

$r2(r2 + 11) - 3(r2 + 11) = 0$

$(r2 - 3)(r2 + 11) = 0$

$since \: radius \: can \: not \: be \: negative$
$so \: r2 = 3cm$

$now \: put \: r2 \: in \: eq2$

$r1 - r2 = 8$

$r1 - 3 = 8$

$r1 = 11cm$

$your \: answer \: is \\ \: r1 = 3cm \: and \: r2 = 11cm$

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