Question

Matlab the equation of a circle with its center at x=3 and y=2 is given by , where r is the radius of the circle. first derive the equation(symbolically) of the tangent line to the circle at the point on the upper part of the circle ( and ). then for specific values of r, , , make a plot of the circle and the tangent line. (r=9, )

Answer 1

The equation of a circle with center at (h,k) and radius r is
$(x-h)^2+(y-k)^2=r^2$
we are given
center is at (3,2)
$(x-3)^2+(y-2)^2=r^2$
solving for y
$(y-2)^2=-1(x-3)^2+r^2$
$y-2=\sqrt{-1(x-3)^2+r^2}$
$y=2+\sqrt{-1(x-3)^2+r^2}$
take the derivitive to find the slope at any point
$\frac{dy}{dx}=((-1)(x-3)^2+r^2))(-x+3)$


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
$y-k=((-1)(x-3)^2+r^2))(-x+3)(x-h)$
where $k=2+\sqrt{-1(h-3)^2+r^2}$
that's the equation of the tangent line at (h,k) for arbitrary values of r

do the plots and other stuff yourself