equation of a tangent line

Question

write the equation of the line that is tangent to x^2+y^2=100; (8,6)

Answer 1

Differentiate wrt x: 2x+2ydy/dx=0. At (8,6) the tangent slope is dy/dx=-x/y=-8/6=-4/3. The tangent line is of the form y=mx+b where m=-4/3 and so that the line passes through (8,6), the tangent point. So 6=(-4/3)*8+b. b=6+32/3=(18+32)/3=50/3 and y=(50-4x)/3 or 3y=50-4x, the equation of the tangent at (8,6).