A circle of finite radius with points $(-2,-2),(1,4),$ and $(k,2006)$ can exist for
$(A)$ no value of $k$
$(B)$exactly one value of $k$
$(C)$exactly two values of $k$
$(D)$infinite values of $k$
Let $(a,b)$ be the center of the circle.
As the three points are on the circumference of the circle.
Then $(a+2)^2+(b+2)^2=(a-1)^2+(b-4)^2=(k-a)^2+(2006-b)^2$
I am stuck here.
Given three arbitrary non-collinear points on a plane, we can always find one and only one circle whose circumference passes through them. So the answer is $(D)$ infinite values of $k$.
Actually, the only value of $k$ for which no such circle exists, is the one making the points collinear, i.e. $$\frac{4-(-2)}{1-(-2)}=\frac{2006-4}{k-1}\Leftrightarrow k=1002$$
P.S.: Given the coplanar but non-collinear points $A,B,C$, we can determine the center of the unique circle coming through them as the point of intersection between the perpendicular bisectors of the line segments $AB$ and $BC$.