Answer:
I'll draw a quarter circle with center O, horizontal radius OB=10. This represents a cross section of the original sphere.
Bisect the vertical radius at point M so that OM = 5.
(This will be oriented on your paper as if it is in the first quadrant.)
Draw a segment from M parallel to the horizontal radius to represent the radius of the small circle; its endpoint is A such that MA=r, presently unknown.
Draw the radius from O to A; (A=10. We now have right triangle OMA.
Observe that its hypotenuse is 10 and its short leg OM=5; therefore MA=r=5√3, either by recognition of the property of 30-60-90 triangles or by use of Pythagorean theorem, if that shortcut were not realized.
