Answer: \(k=\dfrac34\); \(P=\left(-\dfrac45,\dfrac{12}{5}\right)\); normal \(4x+3y=4\).

Substitute \(y=kx+3\) into the curve:

\(x^2-2x+(kx+3)^2=8.\)

Expand:

\(x^2-2x+k^2x^2+6kx+9=8.\)

So

\((1+k^2)x^2+(6k-2)x+1=0.\)

Since the line is a tangent, this quadratic has exactly one root. Therefore its discriminant is zero:

\((6k-2)^2-4(1+k^2)(1)=0.\)

Expand and simplify:

\(36k^2-24k+4-4-4k^2=0.\)

So

\(32k^2-24k=0.\)

Hence

\(8k(4k-3)=0.\)

Since \(k\) is positive,

\(k=\dfrac34.\)

Using \(k=\dfrac34\), the tangent line is

\(y=\dfrac34x+3.\)

At tangency, the repeated root of the quadratic is

\(x=-\dfrac{b}{2a}=-\dfrac{\frac52}{2\cdot\frac{25}{16}}=-\dfrac45.\)

Then

\(y=\dfrac34\left(-\dfrac45\right)+3=\dfrac{12}{5}.\)

So

\(P=\left(-\dfrac45,\dfrac{12}{5}\right).\)

Differentiate the curve implicitly:

\(x^2-2x+y^2=8.\)

This gives

\(2x-2+2y\dfrac{dy}{dx}=0.\)

So

\(\dfrac{dy}{dx}=\dfrac{1-x}{y}.\)

At \(P\), the tangent gradient is \(\dfrac34\), so the normal gradient is

\(-\dfrac43.\)

Use point-gradient form:

\(y-\dfrac{12}{5}=-\dfrac43\left(x+\dfrac45\right).\)

Simplifying gives

\(\boxed{4x+3y=4}.\)