Two orthogonal circles are such that the area | vedclass

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(A) Let the radii of the two circles be $r$ and $r_1$. Given that the area of one is twice the area of the other,we have $\pi r_1^2 = 2 \pi r^2$,which

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$\sqrt{3} r$

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(A) Let the radii of the two circles be $r$ and $r_1$. Given that the area of one is twice the area of the other,we have $\pi r_1^2 = 2 \pi r^2$,which implies $r_1^2 = 2r^2$ or $r_1 = \sqrt{2} r$.For two orthogonal circles with radii $r$ and $r_1$ and distance between centers $d$,the condition for orthogonality is $d^2 = r^2 + r_1^2$.Substituting the values,we get $d^2 = r^2 + (\sqrt{2} r)^2 = r^2 + 2r^2 = 3r^2$.Therefore,the distance between their centers is $d = \sqrt{3} r$.

Two orthogonal circles are such that the area of one is twice the area of the other. If the radius of the smaller circle is $r$,then the distance between their centers will be -

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