Official solution
Given equation of circle
x2 + y2 = 25
\[ \therefore \] Tangent equation at (3, 4)
T : 3x + 4y = 25
Incentre of \[\Delta\]OPQ.
\[I = \left( {{{{{25} \over 4} \times {{25} \over 3}} \over {{{25} \over 3} + {{25} \over 4} + {{125} \over {12}}}},{{{{25} \over 3} \times {{25} \over 4}} \over {{{25} \over 3} + {{25} \over 4} + {{125} \over {12}}}}} \right)\]
\[ \therefore \] \[I = \left( {{{625} \over {75 + 100 + 125}},{{625} \over {75 + 100 + 125}}} \right) = \left( {{{25} \over {12}},{{25} \over {12}}} \right)\]
\[ \because \] Distance from origin to incenter is r.
\[ \therefore \] \[{r^2} = {\left( {{{25} \over {12}}} \right)^2} + {\left( {{{25} \over {12}}} \right)^2} = {{625} \over {72}}\]
Therefore, the correct answer is (B).