The incenter (center of the incircle) of the triangle $ ABC $ is point:
$$ \left(4.7534,~10.1233\right) $$explanation
The incenter of a triangle is given by:
$$ C = \left( ~ \frac{a \cdot A_x + b \cdot B_x + c \cdot C_x}{a + b + c}~,~ \frac{a \cdot A_y + b \cdot B_y + c \cdot C_y}{a + b + c} \right) $$where $ a = d(B,C)$ , $ b = d(A,C) $ and $ c = d(A,B) $.
In this example we have : $ A_x = -\frac{ 5 }{ 3 } $ , $ A_y = \frac{ 245 }{ 9 } $ , $ B_x = 15 $ , $ B_y = 5 $ , $ C_x = 0 $ and $ C_y = 5 $ .
The sides lengths are: $ a = 15 $ , $ b = \frac{ 5 \sqrt{ 1609}}{ 9 } $ and $ c = \frac{ 250 }{ 9 } $.
When we insert these values into the formula, we obtain the given result.