Assume the circle's center is at (0, 0) and, thus, has the simple circle formula of x^2 + y^2 = radius^2, where the x^2 notation is "x squared." Assume that the circle has a radius of 10 and the desired tangent at the x, y coordinates is (6, 8).
Apply the formula to computer the tangent to a curve, which is y - y0 = m(x - x0), where x0, y0 are the coordinates of the tangent point and m is the slope of the tangent line.
Compute "m," which is the slope of the tangent line, by deriving the equation x^2 + y^2 = radius^2. Using the radius = 10, the equation reads x^2 + y^2 = 100. The derivative with respect to x becomes 2x + 2y(dy / dx) = 0. Using algebra to simplify, this becomes 2y(dy / dx) = -2x, then dy / dx = -2x / 2y and finally dy / dx = -x / y. Hence, the final slope is found by inserting the x, y coordinates (6, 8) into the equation and solving to find -6 /8, which simplifies to -3/4.
Apply all the values that have been computed into the tangent formula y - y0 = m(x - x0), so y - 8 = (-3/4)(x - 6), which simplifies to 4y - 32 = -3x + 18. This equation then becomes 4y + 3x = 50, which ultimately reveals that the tangent equation as 3x + 4y - 50 = 0.