Question 1183327
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<pre>

Any straight line, perpendicular to the line 3x - 4y - 4 = 0, has an equation

    4x + 3y = c,


where c is some constant.  The perpendicular line, passing through the point (-4,-4), has the constant "c" equal to 

    4*(-4) + 3*(-4) = -16 + -12 = -28.


So, the circle center is the intersection of the lines

    4x + 3y = -28     (1)

     x +  y =  -7     (2)


It is easy to solve (Elimination or Substitution, on your choice).


The intersection point is  y = 0, x = -7.


Thus the center of the circle is the point (x,y) = (-7,0).


The radius of the circle is the distance from its center (-7,0) to the point (-4,-4)

    r = {{{sqrt((-4-(-7))^2+(-4-0)^2)}}} = {{{sqrt(3^2 + 4^2)}}} = 5.


Thus the center of the circle is the point (-7,0)  and its radius is 5 units.


The standard equation of the circle is


    {{{(x+7)^2}}} + {{{y^2}}} = 25.    <U>ANSWER</U>
</pre>

Solved.