how do I solve for all values of k of a line y=2k tangent to the circle x^2 + y^2 = 5?

Answer 1

y² = 5 - x²

y = sqrt(5 - x²)

dy/dx = -2x/sqrt(5 - x²) = 2k

This takes into account only the + values of the square root; the solution must take the negative value of the square root as well, in order to be complete.

Answer 2

y² = 5 - x²

y = sqrt(5 - x²)

dy/dx = -2x/sqrt(5 - x²) = 2k

This takes into account only the + values of the square root; the solution must take the negative value of the square root as well, in order to be complete.

Answer 3

y² = 5 - x²

y = sqrt(5 - x²)

dy/dx = -2x/sqrt(5 - x²) = 2k

This takes into account only the + values of the square root; the solution must take the negative value of the square root as well, in order to be complete.

Answer 4

y² = 5 - x²

y = sqrt(5 - x²)

dy/dx = -2x/sqrt(5 - x²) = 2k

This takes into account only the + values of the square root; the solution must take the negative value of the square root as well, in order to be complete.

Answer 5

y² = 5 - x²

y = sqrt(5 - x²)

dy/dx = -2x/sqrt(5 - x²) = 2k

This takes into account only the + values of the square root; the solution must take the negative value of the square root as well, in order to be complete.