I have a question about math.?

The center of P travels along a semicircle with a radius of 5, centered on the origin.

If the radius of P is r, then the point on P furthest from the origin describes a semicircle with radius 5+r,

and the point closest to the origin describes a semicircle with radius 5-r. The total area between these semicircles is

π [(5+r)^2 - (5-r)^2]/2

[where the division by 2 is because they're semicircles]

which simplifies:

= π [(25 + 10r + r^2) - (25 -10r + r^2)]/2

= π * 20r / 2

= π * 10r

We come now to the problem of defining the concept "sweeps out." I am going to assume that it's meant to include all the area through which circle P passes temporarily during its travel, INCLUDING the area occupied by circle P at the beginning or end of the process. Because we calculated the area between the semicircles right up to the y-axis, there's a semicircle of P at each end of the path that goes beyond the axis, so we need to add two halves (that is, one whole) area of circle P:

π * 10r + π r^2 = 39π

10r + r^2 = 39 [dividing both sides by π]

r^2 + 10r - 39 = 0

(r + 13) (r - 3) = 0

This has two solutions: r=-13 and r=3. Only the positive solution is useful here:

the radius of circle P is 3 units.