Statement-I     The points $A\left(-6,10\right),B\left(-4,6\right)$  and $C\left(3,-8\right)$  are collinear such that $AB=\frac{2}{9}AC$ .

Statement-II     The point P(−2, 4) lies on a circle of radius 6 and centre (3, 5) .

Statement A:

Given points $A\left(-6,10\right),B\left(-4,6\right)$  and $C\left(3,-8\right)$  are collinear

$AB=\sqrt{{\left(-4+6\right)}^2+{\left(6-10\right)}^2}=\sqrt{2^2+{\left(-4\right)}^2}$

$=\sqrt{4+16}=\sqrt{20}=2\sqrt{5}$

$AC=\sqrt{{\left(3+6\right)}^2+{\left(-8-10\right)}^2}$

$=\sqrt{9^2+{\left(-18\right)}^2}$

$=\sqrt{81+324}=\sqrt{405}=9\sqrt{5}$

$AC=9\times \frac{AB}{2}\Rightarrow AB=\frac{2}{9}AC$  

Statement B:

Given radius of circle $\left(r\right)=6$

Center $\left(A\right)=\left(3,5\right)$

$P=\left(-2,4\right)$

$AP=\sqrt{{\left(-2-3\right)}^2+{\left(4-5\right)}^2}=\sqrt{25+1}=\sqrt{26}$

$r=6=\sqrt{36}>\sqrt{26}$

$r>AP$

$\therefore P$  is not lies on the circle