Question 1065346
The slope of OA is {{{(4-0)/(2-0)=4/2=2}}} .
The slope of AB is {{{(4-6)/(2-(-2))=(-2)/(2+2)=(-2)/4=-1/2}}} .
Since the product of the slopes is
{{{2*(-1/2)=-1}}} , OA and AB are perpendicular.
Angle OAB is an inscribed right angle in the circle we are looking for,
OB is a diameter of that circle,
and the center of the circle is the midpoint of OB,
{{{C(-1,3)}}} .
{{{OB^2=(-2)^2+6^2=4+36=40=diameter^2=(2radius)^2=4*radius^2}}}
So {{{radius^2=40/4=10}}}
With {{{C(-1,3)}}} and {{{radius^2=10}}} ,
we can write the equation for the circle as
{{{(x+1)^2+(y-3)^2=10}}} .