P(x1, y1) is a point on the circle x^2 + y^2 = r^2.. Show that tangent at P has gradient - x1/y1

P(x1, y1) is a point on the circle x^2 + y^2 = r^2
Show that tangent at P has gradient - x1/y1 using implicit differentiation and not using implicit differentiation.

Please help im stuck!

Using implicit:
x^2+y^2=r^2
2x+2y dy/dx=0
dy/dx=-x/y
Plugging in x1 and y1, you get -x1/y1


Not using implicit:
Solve for y
y=(r^2-x^2)^(1/2)
y'=1/2(r^2-x^2)^(-1/2)(-2x)
=-x/((r^2-x^2)^(1/2))
substitute y for the denominator because it is the original equation solved for y
=-x/y
plug in x1 and y1
=-x1/y1

x² + y² = r²
implicit differentiation:
2x + 2yy' = 0
y' = -x/y

Substitute the (x,y) values of any point on the circle to get the slope of the curve at that point. If (x1,y1) is on the circle, then the slope at that point is:
y' = -x/y = -x1/y1