Answer: \(k=7\), and the point of contact is \((1,2)\).

Use gradients for straight-line work: parallel lines have equal gradients and perpendicular lines have gradients whose product is \(-1\).

From the line

\(x+3y=k,\)

we have

\(x=k-3y.\)

Substitute this into the circle equation:

\((k-3y)^2+y^2+2y-9=0.\)

Expanding gives

\(k^2-6ky+9y^2+y^2+2y-9=0.\)

So

\(10y^2+(2-6k)y+(k^2-9)=0.\)

Since the line is a tangent, this quadratic has one repeated root, so its discriminant is zero:

\((2-6k)^2-4(10)(k^2-9)=0.\)

Expanding,

\(4-24k+36k^2-40k^2+360=0.\)

Therefore

\(-4k^2-24k+364=0.\)

Dividing by \(-4\),

\(k^2+6k-91=0.\)

Factorising,

\((k-7)(k+13)=0.\)

Since \(k\) is positive,

\(k=7.\)

Now substitute \(k=7\) into

\(10y^2+(2-6k)y+(k^2-9)=0.\)

This gives

\(10y^2-40y+40=0.\)

Dividing by \(10\),

\(y^2-4y+4=0,\)

so

\((y-2)^2=0.\)

Thus

\(y=2.\)

Using \(x+3y=7\),

\(x+6=7,\)

so

\(x=1.\)

The point of contact is therefore

\((1,2).\)