Show that the following four points in each of the following are concyclic and find the circle on which they lie
$(1,1),(-6,0),(-2,2),(-2,-8)$
$(1,2),(3,-4),(5,-6),(19,8)$
$(1,-6),(5,2),(7,0),(-1,-4)$
$(9,1),(7,9),(-1,12),(6,10)$

Fore concyclic

$BD\times AC=AD.BC+AB.DC$

(1) $A(1,1),B(-6,0),C(-2,2),D(-2,-8)$

$\Rightarrow \sqrt{(-6+2{)}^2+(8{)}^2}.\sqrt{3^2+1^2}=\sqrt{3^2+9^2}.\sqrt{4^2+2^2}+\sqrt{7^2+1}.\sqrt{10^2+10^2}$

$\Rightarrow \sqrt{16+64}.\sqrt{10}=\sqrt{1800}+10\sqrt{50}$

$\Rightarrow 20\sqrt{2}=20\sqrt{2}$

Hence they are concyclic

(2) $(1,-2),(3,4)(5,-6),(19,8)$

$\Rightarrow \sqrt{16^2+12^2}.\sqrt{4^2+8^2}=\sqrt{18^2+6^2}.\sqrt{2^2+6^2}+\sqrt{2^2+6^2}\sqrt{14^2+14^2}$

$\Rightarrow \sqrt{400}.\sqrt{80}=\sqrt{360}.\sqrt{8}+\sqrt{40}\sqrt{392}$

$\Rightarrow 80\sqrt{5}=80\sqrt{5}$

Hence there are concyclic.

(3) $(1,-6),(5,2),(7,0),(-1,-4)$

$\Rightarrow \sqrt{6^2+6^2}.\sqrt{6^2+6^2}=\sqrt{2^2+2^2}.\sqrt{2^2+2^2}+\sqrt{4^2+8^2}.\sqrt{8^2+4^2}$

$\Rightarrow \sqrt{36+36}.\sqrt{36+36}=\sqrt{8}.\sqrt{8}+\sqrt{16+64}+\sqrt{16+64}$

$\Rightarrow \sqrt{72}.\sqrt{72}=80+8$

$72=88$

Hence they are not concyclic

(4) $(9,1),(7,9),(-1,12),(6,10)$

$\Rightarrow \sqrt{(1{)}^2+(1{)}^2}.\sqrt{10^2+11^2}=\sqrt{3^2+9^2}.\sqrt{8^2+3^2}+\sqrt{2^2+8^2}\sqrt{7^2+20^2}$

$\Rightarrow \sqrt{2}.\sqrt{221}=\sqrt{90}\sqrt{73}+\sqrt{68}\sqrt{53}$

$\Rightarrow \sqrt{442}=\sqrt{6570}+\sqrt{3604}$

$\Rightarrow \sqrt{442}=\sqrt{6629}$

Hence they are no concyclic