Find the equation of the tangents to the circle x^2+y^2=9 with a slope=1

x^2 + y^2 = 9

Differentiate wrt x

2x + 2y*dy/dx = 0

Hence x + y = 0. (because dy/dx = 1), so y = -x

Put this in the original equation to get. 2x^2 = 9

Two solutions:  x = 3/sqrt(2) and x = -3/sqrt(2)

Corresponding values of y are y = -3/sqrt(2) and y = 3/sqrt(2) (because y = -x)

Equation of line with slope 1 is y = x + k where k is a constant.

Plug the known pairs of x and y into this equation to find the corresponding values of k.

Find the equation of the tangents to the circle x^2+y^2=9 with a slope=1

Implicitly differentiate

2x + 2yy'  = 0      divide through by 2

x + yy'  = 0

yy'  = -x

y'  = -x / y

If the slope is 1, then  y = -x       and we can use the equation to find the points where this occurs

x^2 + (-x)^2  = 9

2x^2  = 9

x^2  = 9/2

x = ± 3 / √2    so.....when  x = 3 / √2, y = -x =  - 3 / √2       and when x = -3 / √2, y = -x =  3 / √2

So.....we have two tangent line equations....the first is

y = (x - 3 / √2) - 3 / √2  →    y = x - 6 / √2 

And the second is

y = (x + 3/ √2 )   + 3 / √2    →   y = x + 6 / √2 

Here's a graph : https://www.desmos.com/calculator/vszg67cvt8

I have included the line y = x which also has a slope of 1 to show that our  two lines are parallel to this one... thus, all three lines have a slope of 1....!!!