At what point is the slope of tangent line -1?

Question

I'm given x^2+y^2=4

Answer 1

The circle equation is x^2+y^2=4 .

To find the tangent of the function , equate the first derivative to zero .

2x + 2y y' = 0

2y y' = -2x

y' = -x/y

But the given slope is -1 equate it to y' .

 y' = -x/y = -1

-x/y = -1

x/y = 1

x = y

Now put x = y in circle equation to find the point where the tangent located .

x^2+y^2=4

x^2+x^2=4 

2x² = 4

x² = 2

x =  √2 

So y =  √2 

So the circle has a tangent at ( √2  ,  √2  ) .