Find the equation of the line tangent to the circle.?

Question

Find the equation of the line tangent to the circle x^2+y^2=28 at the point (3 sqrt 2, 3)

Answer 1

2x + 2y dy/dx = 0
dy/dx = - 2x / 2y
dy/dx = - x / y
dy/dx = - 3√2 / 3 = - √2
y - 3√2 = (-√2) (x - 3)
y = (-√2) x + 3√2 + 3√2
y = (-√2) x + 6√2

Answer 2

First, find the slope m by applying the rules of implicit differentiation:

d[(x ^ 2) + (y ^ 2)] / dt = d[28] / dt

2 * x + 2 * y * (dy / dt) = 0.

Solve for dy/dt, which is the slope m of the tangent line to the given equation of a circle:

dy / dt = m = -sqrt(2).

Next, use the point-slope formula and substitute the given x and y values for x_1 and y_1; solve for y to obtain the slope - intercept form of the line tangent to the equation of the given circle:

y - y_1 = m * (x - x_1) = y - 3 = -sqrt(2) * {x - [3 * sqrt(2)]} => y = [- sqrt(2) * x] - [3 * sqrt(2) * sqrt(2)] +3 = [- sqrt(2) * x] - 3.

Answer 3

(3 sqrt2, 3) is not on the circle. If you were to plug those numbers into the equation it equals only 27. It is actually a point inside the circle. You might want to reread the question.