Question 1183327
Let {{{(x-x[0])^2+(y-y[0])^2 = d^2}}} be the equation of the circle.  

Note that the point ({{{x[0]}}}, {{{y[0]}}}) is on the line x + y + 7 = 0, hence {{{x[0] + y[0] + 7 = 0}}}, or {{{ y[0] = -x[0] - 7}}}.

Since the point (-4,-4) is on the circle, then the the equation  

{{{(-4-x[0])^2+(-4-y[0])^2 = d^2}}} is true.  This becomes {{{(x[0]+4)^2+(y[0]+4)^2 = d^2}}} after simplifying.

Also, {{{(x[0]+4)^2+(-x[0] - 7 +4)^2 = d^2}}}, or {{{(x[0]+4)^2+(x[0] +3)^2 = d^2}}} after the fact {{{ y[0] = -x[0] - 7}}}.

 {{{(x[0]+4)^2+(x[0] +3)^2 = d^2}}} <==> {{{2x[0]^2+14x[0]+25 = d^2}}}.


Since the point (-4,-4) is on the line 3x - 4y - 4 = 0, and the radial line is perpendicular to this line, 
we can find the distance between 3x - 4y - 4 = 0 and the center ({{{x[0]}}}, {{{y[0]}}}).  
The distance is given by {{{d = abs(3x[0] - 4y[0]-4)/sqrt(3^2+4^2) = abs(3x[0] - 4y[0]-4)/5 = abs(3x[0] - 4(-x[0]-7)-4)/sqrt(3^2+4^2) = abs(7x[0]+24)/5}}}.

==> {{{d^2 = (7x[0]+24)^2/25 = (49x[0]^2 + 336x[0] + 576)/25}}}


==>  {{{2x[0]^2+14x[0]+25 = (49x[0]^2 + 336x[0] + 576)/25}}}, or

{{{50x[0]^2 +350x[0]+ 625 = 49x[0]^2 + 336x[0] + 576}}}, or {{{x[0]^2 + 14x[0] +49 = 0}}}, or {{{(x[0] + 7)^2 = 0}}} ==> {{{x[0] = -7}}}.

Then {{{ y[0] = -x[0] - 7}}} ==> {{{y[0] = 0}}}. Therefore the center is (-7,0).

Also, {{{d^2 = (7x[0]+24)^2/25 = (7*-7+24)^2/25 = 25}}}.

Therefore the equation of the circle is {{{highlight((x+7)^2+y^2 = 25)}}}.

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This is practically an engineering problem, and some other tutor here might have a much shorter solution. 
(Admittedly I'm working from memory on this one.) 
From what I remember there is an even shorter formula for a problem like this, but I cannot retrieve that formula.