8,9

Given a regular tetrahedron \( PABC \) with edge length \( a \). A sphere is drawn through the points \( C, E, M, P \), where \( E \) is the midpoint of \( AB \), and \( M \) is the midpoint of \( AC \). Find the radius of the sphere.
The center $O$ of the given sphere of radius $R$ is equidistant from points $E, M$, and $C$, so it lies on the perpendicular $l$ to the plane $E M C$ (i.e., to the base plane $A B C$ of the tetrahedron), passing through the center of the circle of triangle $E M C$. If $Q$ is the midpoint of side $B C$ of the equilateral triangle $A B C$, then $Q E=Q M=Q C$, i.e., $Q$ is the center of the circle passing through points $E, M$, and $C$. Let $P H$ be the height of the tetrahedron $A B C D$. Then $P H \| O Q$. Through the parallel lines $P H$ and $O Q$, we draw a plane. Let $O Q=x$. Drop a perpendicular $O T$ from the vertex $O$ of the right trapezoid $O Q H P$ to the base $P H$. By the Pythagorean theorem in the right triangle $O T P$, we find that

$$
R^2=O P^2=O T^2+T P^2=Q H^2+(P H-T H)^2=
$$

$$
=Q H^2+(P H-O Q)^2=\left(\frac{a \sqrt{3}}{6}\right)^2+\left(a \sqrt{\frac{2}{3}}-x\right)^2
$$

and from the right triangle $O Q C-$

$$
R^2=O C^2=O Q^2+Q C^2=x^2+\left(\frac{a}{2}\right)^2=x^2+\frac{a^{2}}{4}
$$

From the equation

$$
\left(\frac{a \sqrt{3}}{6}\right)^2+\left(a \sqrt{\frac{2}{3}}-x\right)^2=x^2+\frac{a^{2}}{4}
$$

we find that $x=\frac{a \sqrt{6}}{8}$. Therefore,

$$
R=\sqrt{x^{2}+\frac{a^{2}}{4}}=\sqrt{\left(\frac{a \sqrt{6}}{8}\right)^{2}+\frac{a^{2}}{4}}=\frac{a \sqrt{22}}{8}
$$

## Answer

$\frac{a \sqrt{22}}{8}$