Question 936969
If it's tangent to the line, then a line from the center point to the intersection point is perpendicular to the tangent line.
{{{2x+7y=10}}}
{{{7y=-2x+10}}}
{{{y=-(2/7)x+10/7}}}
The perpendicular line would have a slope of 
{{{m(-2/7)=-1}}}
{{{m=7/2}}} Slope of the perpendicular line
since perpendicular lines have slopes that are negative reciprocals.
Using the point,
{{{y-(-5)=(7/2)(x-(-4))}}}
{{{y+5=(7/2)(x+4)}}}
{{{y+5=(7/2)x+14}}}
{{{y=(7/2)x+9}}} Equation of the perpendicular line
Find the intersection point of the two lines.
{{{(7/2)x+9=-(2/7)x+10/7}}}
Multiply both sides by 14.
{{{49x+126=-4x+20}}}
{{{53x=-106}}}
{{{x=-2}}}
{{{y=(7/2)(-2)+9}}}
{{{y=-7+9}}}
{{{y=2}}}
The distance from (-4,-5) to (-2,2) is the radius of the circle.
{{{R^2=(-2-(-4))^2+(2-(-5))^2}}}
{{{R^2=(-2+4)^2+(2+5)^2}}}
{{{R^2=2^2+7^2}}}
{{{R^2=4+49}}}
{{{R^2=53}}}
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{{{highlight_green((x+4)^2+(y+5)^2=53)}}}
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{{{drawing(300,300,-8,8,-8,8,grid(1),circle(-4,-5,0.2),circle(-2,2,0.2),circle(-4,-5,sqrt(53)),graph(300,300,-8,8,-8,8,(-2x+10)/7))}}}