for what value of K will the line with equation y=kx be tangent to the circle with equation ( x-7)^2 + (y-6)^2 =9 ?
Taking the equation of the circle and then taking the derivative, we get 2(x - 7) + 2(y - 6)*dy/dx = 0, so dy/dx = -(x - 7)/(y - 6).
We can also parameterize the circle by taking x = 7 + cos(t), y = 6 + sin(t), so dy/dx = -cos(t)/sin(t) = k.
Since the tangent also goes through the origin, k = (6 + sin(t))/(7 + cos(t)), so (6 + sin(t))/(7 + cos(t)) = -cos(t)/sin(t). The solutions (cos(t),sin(t)) are (-0.729305, 0.684189) and (0.564599,-0.25365), so the possible values of k are 1.06594 and 0.684059.