Question 1139514
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Put the equations in slope-intercept form.<br>
{{{4x+3y=4}}} -->  {{{y = -(4/3)x+4/3}}}
{{{x-y=7}}} -->  {{{y = x-7}}}<br>
If the circle is tangent to the line {{{y = -(4/3)x+4/3}}} at (4,-4), then the center of the circle has to lie on the line with slope 3/4 passing through (4,-4).  (A radius to a point of tangency is perpendicular to the tangent; slopes of perpendicular lines are negative reciprocals.)<br>
{{{y = (3/4)x+b}}}
{{{-4 = (3/4)4+b}}}
{{{-4 = 3+b}}}
{{{b = -7}}}<br>
So the center of the circle lies on the line {{{y = (3/4)x-7}}}.<br>
The given information says that the center of the circle lies on the line {{{y = x-7}}}.<br>
So the center of the circle is at the one point that lies on both {{{y = (3/4)x-7}}} and {{{y = x-7}}}.<br>
Solve that pair of equations to find the intersection point; that is the center of the circle.