Question 201095
Try this:
Original area is: 
{{{A[1] = pi*r^2}}}
The area of the new circle is:
{{{A[2] = 2*A[1]}}} and {{{A[2] = pi*(r+4)^2}}}, so...
{{{pi*(r+4)^2 = 2*pi*r^2}}} 
{{{pi*(r^2+8r+16) = 2*pi*r^2}}} Divide both sides by {{{pi}}}
{{{r^2+8r+16 = 2*r^2}}} Subtract {{{r^2}}} from both sides.
{{{8r+16 = r^2}}} Rewrite as a quadratic equation in standard form:
{{{r^2-8r-16=0}}} Solve using the quadratic formula: {{{r = (-b+-sqrt(b^2-4ac))/2a}}} where: a = 1, b = -8, and c = -16.
{{{r = (-(-8)+-sqrt((-8)^2-4(1)(-16)))/2(1)}}} Simplify:
{{{r = (8+-sqrt(64-(-64)))/2}}}
{{{r = (8+-sqrt(128))/2}}}
{{{r = 4+4sqrt(2)}}} or {{{r = 4-4sqrt(2)}}} or approximately...
{{{highlight_green(r = 9.65685)}}} or {{{cross(r = -1.65685)}}} Discard the negative solution as the radius can only be a positive value!