Question 270996
We have a circle with radius 10 centered at (0,0) and passing through (8,6).
The slope of (0,0) and (8,6) is 
{{{m = 3/4}}}
but we want the perpendicular slope for the tangent line as
{{{ m = _4/3}}}
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The equation is expressed as
{{{y = mx + b}}}
and putting in our slope, -4/3 and point, (8,6) we get
{{{6 = (-4/3)*8 + b}}}
which is
{{{6 = -32/3 + b}}}
and then
{{{b = 50/3}}}
The answer is
{{{y = (-4/3)x + 50/3}}}