A(4, –2), B(–2, 4) and C(5, 5) are the vertices of a $∆$ABC. If the perpendicular bisector of BC meets the circumcircle at D. A and D are on the opposite side of BC. Then

$Given vertices of triangle are A=(4,-2),B=(-2,4),C=(5,5)$

For this triangle circum centre =$\left(\frac{15}{8},\frac{15}{8}\right)$
Incentre =$\left(\frac{5}{2},\frac{5}{2}\right)$

AD must pass through the incentre of the triangle.
and D$=\left(\frac{5}{4},\frac{25}{4}\right)$

Equation of AD is

 $$\begin{gathered} y+2=\frac{{\displaystyle \frac{25}{4}}+2}{{\displaystyle \frac{5}{4}}-4}(x-4) \\ \Rightarrow 3x+y-10=0 \\ Dis\tan ce from (0,0) to AD is \sqrt{10} \end{gathered}$$