Equal parts

Let $r$ be the radius of the circle, $R$ be the area of the red region and $Y$ be the area of the yellow region. Then

$\cos 30=\dfrac{AB}{10}$

$\dfrac{\sqrt{3}}{2}=\dfrac{AB}{10}$

$AB=5\sqrt{3}$

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$R=area~of~a~sector=\dfrac{30}{360}\pi r^2=\dfrac{\pi r^2}{12}$

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$Y=area~of~triangle~ABC-area~of~a~sector=\dfrac{1}{2}(10)(5\sqrt{3})(\sin 30)-\dfrac{\pi r^2}{12}=\dfrac{1}{2}(10)(5\sqrt{3})(\dfrac{1}{2})-\dfrac{\pi r^2}{12}=\dfrac{25}{2}\sqrt{3}-\dfrac{\pi r^2}{12}$

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$R=Y$

$\dfrac{\pi r^2}{12}=\dfrac{25}{2}\sqrt{3}-\dfrac{\pi r^2}{12}$

$r^2=\dfrac{300\sqrt{3}}{4\pi}$

$r=\sqrt{\dfrac{100*3\sqrt{3}}{4\pi}}=\dfrac{10}{2}\sqrt{\dfrac{3\sqrt{3}}{\pi}}=5\sqrt{\dfrac{3\sqrt{3}}{\pi}}$

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$a+b=5+3=\boxed{8}$