Question 938729
Circle: {{{(x+4)^2 + (y-2)^2 = r^2}}}
tangent to...
y = 2x + 2
...
Line perpendicular thru C(-4,2)
y - 2 = (-1/2)(x + 4)
y = (-1/2)x
...
(-1/2)x = 2x + 2
-2 = (5/2)x
(-4/5) = x  and y = 2/5 {{{(-1/2)(-4/5)}}}
...
C(-4,2)
(-4/5, 2/5) 
r^2 = {{{ (x[1]-x[2])^2+ (y[1]-y[2])^2  }}}
...
r^2 = {{{(-16/5)^2 + (8/5)^2}}} = 320/25
{{{(x+4)^2 + (y-2)^2 = r^2}}}
{{{drawing(300,300,    -10,10,-10,10,   
 grid(1),
circle(-4,2,0.4),
circle(-.8, .4,.4),
graph( 300, 300, -10,10,-10,10,0, 2x+2, -.5x))}}}