the equation of a circle with center at (h,k) and radius r is
[tex](x-h)^2+(y-k)^2=r^2[/tex]
we are given
center is at (3,2)
[tex](x-3)^2+(y-2)^2=r^2[/tex]
solving for y
[tex](y-2)^2=-1(x-3)^2+r^2[/tex]
[tex]y-2=\sqrt{-1(x-3)^2+r^2}[/tex]
[tex]y=2+\sqrt{-1(x-3)^2+r^2}[/tex]
take the derivitive to find the slope at any point
[tex]\frac{dy}{dx}=((-1)(x-3)^2+r^2))(-x+3)[/tex]
we can use point slope form
for apoint (h,k) and slope m, the equation is
y-k=m(x-h)
not sure if you want in terms of what
I'll just say for a point (h,k)
so we get
[tex]y-k=((-1)(x-3)^2+r^2))(-x+3)(x-h)[/tex]
where [tex]k=2+\sqrt{-1(h-3)^2+r^2}[/tex]
that's the equation of the tangent line at (h,k) for arbitrary values of r
do the plots and other stuff yourself
