What's the most efficient way to draw this region?

f1h = HoldForm[10 - Sqrt[10^2 - x^2]];
f1 = f1h // ReleaseHold;
f2h = HoldForm[5 - Sqrt[5^2 - (x - 5)^2]];
f2 = f2h // ReleaseHold;
f3h = HoldForm[5 + Sqrt[5^2 - (x - 5)^2]];
f3 = f3h // ReleaseHold;

The x values for the curve intersections are

{x1, x2} = x /. Solve[f1 == #, x][[1]] & /@ {f2, f3};

The point coordinates for the curve intersections are

{pt1, pt2} = ({#, f1 /. x -> #} // FullSimplify) & /@ {x1, x2};

reg1 = ImplicitRegion[f1 < y < 10 && x > 0, {x, y}];
reg2 = ImplicitRegion[f2 < y < f3, {x, y}];

Show[
 Region[
  regDiff = RegionDifference[reg2, reg1]],
 Plot[{f1, f2, f3}, {x, 0, 10},
  PlotStyle -> {
    {Orange, AbsoluteThickness[4]},
    {Purple, AbsoluteThickness[4]},
    {Darker@Green, AbsoluteThickness[4]}}],
 Epilog -> {
   Text[Style[x1, 14, Bold], {x1, 2}, {0, -1}],
   Arrow[{pt1, {x1, 2}}],
   Text[Style[x2, 14, Bold], {7.75, pt2[[2]]}, {1, 0}],
   Arrow[{pt2, {7.75, pt2[[2]]}}],
   Text[Style[f1h, 14, Bold], {10, 8}, {-1, 0}],
   Text[Style[f2h, 14, Bold], {9.5, 1.5}, {-1, 0}],
   Text[Style[f3h, 14, Bold], {5, 11}],
   Red, AbsolutePointSize[7],
   Point[{pt1, pt2}]},
 Ticks -> {{5, 10}, {5, 10}},
 PlotRange -> {{-1, 14}, {-1, 12}},
 Axes -> True]

The area of the shaded region is

area = Area[regDiff] // FullSimplify

(* 25/2 (Sqrt[7] + π - ArcCot[3/Sqrt[7]] - 4 ArcTan[(5 Sqrt[7])/9]) *)

The area relative to the smaller circle is

area/Area[reg2] // N

(* 0.186378 *)

One simple way to visualize complicated regions in mathematica

disk1 = Region[Disk[{5, 5}, 5]]

disk2 = Region[Disk[{0, 10}, 10]]

disk3 = Region[Disk[{10, 0}, 10]]

result = Region[
  RegionUnion[RegionDifference[disk1, disk3], 
   RegionDifference[disk1, disk2]]]

RegionPlot[
  x^2 + y^2 < 25 \[And] 
 ((x - 5)^2 + (y + 5)^2 > 100 \[Or] (x + 5)^2 + (y - 5)^2 > 100),
 {x, -5, 5}, {y, -5, 5}]

Or...

z[w_] := EuclideanDistance[{x, y}, w {5, -5}];
RegionPlot[
 z[0] < 5 \[And] (z[1] > 10 \[Or] z[-1] > 10),
 {x, -5, 5}, {y, -5, 5}]

Or...

z[w_] := (a = ({x, y} - 5 {w, -w})).a;
RegionPlot[
 z[0] < 25 \[And] (z[1] > 100 \[Or] z[-1] > 100), 
 {x, -5, 5}, {y, -5, 5}]

Or even shorter....

z[w_, k_] := (a = ({x, y} - 5 {w, -w})).a > 25 k;
RegionPlot[
Not[z[0, 1]] \[And] (z[1, 4] \[Or] z[-1, 4]), 
{x, -5, 5}, {y, -5, 5}]

I would be very impressed if someone uses fewer keystrokes than this in a Region-based solution:

d[c_, r_:10] := Region[Disk[c, r]];
RegionDifference[d[{5, 5}, 5], RegionIntersection[d[{0, 10}], d[{10, 0}]]]

Or...

d[c_, r_:10] := Region[Disk[c, r]];
q = {0, 10};
h = {5, -5};
RegionDifference[d[q + h, 5], RegionIntersection[d[q], d[q + 2 h]]]

Or....

d[c_, r_:10] := Region[Disk[c, r]]; q = {5, 5}; h = {5, -5};
RegionDifference[d[q, 5], RegionIntersection[d[q - h], d[q + h]]]

Pretty efficient!