Answer: \(P=(3,3)\), and the required circle is \((x-3)^2+(y-3)^2=18\).

First complete the square:

\(x^2-10x+y^2-4y+24=0,\)

\((x-5)^2+(y-2)^2=5.\)

So the centre is \(C=(5,2)\) and the radius is \(\sqrt5\).

The tangent \(y=2x-3\) has gradient \(2\), so the radius to the point of contact has gradient \(-\frac12\).

The radius through \(C=(5,2)\) has equation

\(y-2=-\frac12(x-5).\)

Substitute \(y=2x-3\):

\(2x-3-2=-\frac12(x-5).\)

So

\(2x-5=-\frac12x+\frac52.\)

Multiplying by \(2\),

\(4x-10=-x+5,\)

hence \(5x=15\), so \(x=3\). Then \(y=2(3)-3=3\). Therefore

\(P=(3,3).\)

The distance from \(P\) to the origin is

\(\sqrt{3^2+3^2}=\sqrt{18}.\)

Thus the required circle is

\((x-3)^2+(y-3)^2=18.\)