# Given the relation $x^2-4x+y^2-4y+4=0$ determine the following… verify answers

a) The equation of the secant line through the intercepts.

For this part I got $y=-x+2$ as the equation of the secant line

b) The slope of the curve at any point on the curve.

Here I got $\frac{dy}{dx}=-\frac{x-2}{y-2}$

c) The equation of the tanget to the curve where $x=4$

Here's where I'm stuck, because when $x=4, y=2$ and if you plug in 2 for $y$ in the previous formula then you're dividing by 0.

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The tangent line is vertical. – André Nicolas May 2 '13 at 5:00

The calculations are fine. As you point out, $\frac{dy}{dx}$ does not exist at $y=2$. Sometimes one says the slope is infinite there.
It may be useful to complete the square, getting $(x-2)^2+(y-2)^2=4$. This is a circle centre $(2,2)$, radius $2$. The tangent line at $x=4$ is vertical, so has equation $x=4$.