Найдём координаты векторов \(\overrightarrow {CA} \) и \(\overrightarrow {CB} \):
\(\overrightarrow {CA} = \left( {2-x;2-0} \right) = \left( {2-x;2} \right);\,\,\,\,\,\,\,\,\overrightarrow {CB} = \left( {6-x;10-0} \right) = \left( {6-x;10} \right).\)
Длина вектора \(\vec d\left( {x;y} \right)\) равна: \(\left| {\vec d} \right| = \sqrt {{x^2} + {y^2}.} \)
Тогда: \(\left| {\overrightarrow {CA} } \right| = \sqrt {{{\left( {2-x} \right)}^2} + 4} ;\,\,\,\,\,\,\,\,\left| {\overrightarrow {CB} } \right| = \sqrt {{{\left( {6-x} \right)}^2} + 100} .\)
Так как по условию \(CA = CB\), то:
\(\sqrt {{{\left( {2-x} \right)}^2} + 4} = \sqrt {{{\left( {6-x} \right)}^2} + 100} \,\,\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\,\,\,4-4x + {x^2} + 4 = 36-12x + {x^2} + 100\,\,\,\,\,\,\, \Leftrightarrow \)
\( \Leftrightarrow \,\,\,\,\,\,\,\,\,\,8x = 128\,\,\,\,\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\,\,\,x = 16.\)
Ответ: 16.