Here is a pictre that shows the problem:
What would be the ratio of the red to blue, and if these were then made spheres how could one find the portion of the surface area that would be inside the other sphere.
Edit: I have found the red to be 1/3 of the circumference for circles, but I'm still not sure how to go about the question for spheres.
Regarding the ratio in $d$ dimensions, note that the hyperspheres always intersect halfway between the centres, so we're looking for the ratio of the area of a hyperspherical cap of the unit sphere of height $1/2$ to the area of the entire sphere, which can be viewed as a hyperspherical cap of height $2$. Thus, the numerical factors in the formula Rahul linked to cancel, and we're just looking for the ratio
$$ \lambda_d=\frac{\displaystyle\int_{\pi/6}^{\pi/2}\cos^{d-2}\phi\,\mathrm d\phi}{\displaystyle\int_{-\pi/2}^{\pi/2}\cos^{d-2}\phi\,\mathrm d\phi}\;. $$
Using Wolfram|Alpha to evaluate the integrals, I get
$$ \begin{align} \lambda_2&=\frac{\pi/3}{\pi}=\frac13\approx0.333\;,\\ \lambda_3&=\frac{1/2}{2}=\frac14=0.250\;,\\ \lambda_4&=\frac{\pi/6-\sqrt3/8}{\pi/2}=\frac13-\frac{\sqrt3}{4\pi}\approx0.196\;,\\ \lambda_5&=\frac{5/24}{4/3}=\frac5{32}\approx0.156\;,\\ \lambda_6&=\frac{\pi/8-9\sqrt3/64}{3\pi/8}=\frac13-\frac{3\sqrt3}{8\pi}\approx0.127\;. \end{align} $$
We can also obtain an asymptotic estimate for the ratio. In the numerator, for large $d$ the values near $\phi=\pi/6$ will dominate, and we can approximate the integrand by a decaying exponential:
$$ \begin{align} \int_{\pi/6}^{\pi/2}\cos^{d-2}\phi\,\mathrm d\phi &= \int_{\pi/6}^{\pi/2}\mathrm e^{(d-2)\log\cos\phi}\,\mathrm d\phi \\ &\approx \int_0^\infty\mathrm e^{(d-2)(\log\cos\pi/6-\phi\tan\pi/6)}\,\mathrm d\phi \\ &= \left(\frac{\sqrt3}2\right)^{d-2}\frac{\sqrt3}{d-2}\;. \end{align} $$
In the denominator, for large $d$ the values near $\phi=0$ will dominate, and we can approximate the integrand by a Gaussian:
$$ \begin{align} \int_{-\pi/2}^{\pi/2}\cos^{d-2}\phi\,\mathrm d\phi &= \int_{-\pi/2}^{\pi/2}\mathrm e^{(d-2)\log\cos\phi}\,\mathrm d\phi \\ &\approx \int_{-\infty}^\infty\mathrm e^{-(d-2)\phi^2/2}\,\mathrm d\phi \\ &= \sqrt{\frac{2\pi}{d-2}}\;. \end{align} $$
Thus, for large $d$ the ratio is approximately given by
$$ \lambda_n\approx\left(\frac{\sqrt3}2\right)^{d-1}\sqrt{\frac2{(d-2)\pi}}\;. $$
For instance, for $d=26$ we have
$$ \lambda_{26}=\frac{\displaystyle\int_{\pi/6}^{\pi/2}\cos^{24}\phi\,\mathrm d\phi}{\displaystyle\int_{-\pi/2}^{\pi/2}\cos^{24}\phi\,\mathrm d\phi}\approx\frac{0.002004}{0.5064}\approx0.00396\;, $$
and the approximation yields
$$ \lambda_{26}\approx\left(\frac{\sqrt3}2\right)^{25}\sqrt{\frac2{24\pi}}\approx0.00447\;. $$
Let $O_1$ and $O_2$ be the centres of the circles, and let $P$ be one of the points of intersection of the circles. Then $\overline{O_1P},\overline{O_2P}$, and $\overline{O_1O_2}$ are all radii of one or the other circle, and $\triangle O_1O_2P$ is equilateral. Thus, $\angle O_2O_1P=\frac{\pi}3$ (or $60^\circ$), and the red arc subtends an angle of $\frac{2\pi}3$ (or $120^\circ)$ at the centre $O_2$ of the righthand circle. In other words, it’s one-third of the circle, so the blue arc is the other two-thirds, and the ratio of red to blue is $1:2$.
For spheres imagine rotating the picture about a horizontal axis running through the centres of the circles. The red line will generate a sort of ‘polar cap’, if you imagine the north pole of of the righthand circle as being at $O_1$, the centre of the lefthand circle. Finding the area of this cap is a fairly straightforward second-semester calculus problem in calculating areas of surfaces of revolution. If you solve that problem in general, not just for polar caps but for the surface area between any two circles of latitude, you’ll make the following surprising (and useful) discovery:
Let $S$ be a sphere, and let $P_1$ and $P_2$ be parallel planes that both intersect $S$. Then the area of the part of $S$ lying between $P_1$ and $P_2$ depends only on the perpendicular distance between $P_1$ and $P_2$ and not on where they intersect $S$.
Applying this to your problem, we see that if $r$ is the radius of the spheres, then we can take $P_1$ to be tangent to the righthand sphere at the pole $O_1$, so that $P_2$, passing through the circle of latitude along which the spheres intersect, is $\frac{r}2$ units away from $P_1$. Let $P_3,P_4$, and $P_5$ be three more planes parallel to $P_1$ and $P_2$ and spaced $\frac{r}2$ units apart. ($P_3$ passes through $O_2$, the centre of the sphere, and $P_5$ is tangent to the sphere at the other pole.) The separations between adjacent planes are all the same, so they slice the surface of the sphere into two caps and two bands that are all of equal area. Clearly that area must be a quarter of the surface area of the sphere.
This MathWorld entry states that useful result a little differently and has a helpful picture.