Question 1115770
Find the equation of circle  center at (1,3) and tangent to the line 5x-12y=8.
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You can find the equation of the line perpendicular to the given line thru the point.
then find the distance from the point to the line, that's the radius.
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Then {{{(x-1)^2 + (y-3)^2 = r^2}}} is the circle.
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Or, use
5x-12y-8 = 0
a = 5, b = -12, c = -8
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{{{r = ABS(a*x[0] + b*y[0] + c)/sqrt(a^2+b^2)}}} to find the distance.
{{{(x[0] y[0])}}} is the point.