What is the equation of the line tangent to x^2+y^2=2 at the point (1,1)
The equation of the circle is x^2+y^2 = 2. The value of the derivative `dy/dx` at at point gives the slope of the tangent at that point.
`(d(x^2+ y^2))/dx = 0`
=> `2x + 2y*(dy/dx) = 0`
=> `dy/dx = -x/y`
At the point (1,1) the slope of the tangent is -1.
The equation of the tangent is `(y - 1)/(x - 1) = -1`
=> y - 1 = 1 - x
=> x + y - 2 = 0
The required equation of the tangent is x + y - 2 = 0