Stuck... beginner

Hello, yakkow!

Quote:

Find the normal to the curve $x^2 + y^2 = 2x + 2y$ that is parallel to the line $x+y = 0$

Differentiate implicitly: . $2x + 2y\cdot y' \:= \:2 + 2y' \quad\Rightarrow\quad 2y\cdot y' - 2y'\;=\;2 - 2x$

Factor: . $2(y - 1)y'\:=\:2(1 - x)\quad\Rightarrow\quad y'\:=\:\frac{1-x}{y-1}$ . . . slope of the tangent

The slope of the normal is: . $m\:=\:\frac{y-1}{x-1}$

The line $x + y\:=\;0\quad\Rightarrow\quad y \,=\,-x$ has slope $-1.$

So we have: . $\frac{y - 1}{x - 1}\:=\:-1\quad\Rightarrow\quad y\,=\,2 - x$

Substitute into the original equation:
. . . $x^2 + (2 - x)^2\:=\:2x + 2(2 - x)\quad\Rightarrow\quad2x^2 - 4x\:=$ $0\quad\Rightarrow\quad x \,=\,0,\;2$

The corresponding y-values are: $y\,=\,2,\;0$

We seem to have two normals: one at (2,0), another at (0,2)

At (2,0) with slope -1, the equation of the normal is:
. . . $y - 0\:=\:-1(x - 2)\quad\Rightarrow\quad y \:=\:-x + 2$

At (0,2) with slope -1, the equation of the normal is:
. . . $y - 2\:=\:-1(x - 0)\quad\Rightarrow\quad y\:=\:-x + 2$

Ha! . . . It's the same line!