The greatest distance of the point $P(10,7)$  from the circle $x^2+y^2-4x-2y-20=0$ is

$S=x^2+y^2-4x-2y-20$ given point $P(10,7)$

$\begin{array}{l}{S}_1={10}^2+7^2-4\left(10\right)-2\left(7\right)-20\\ =100+49-40-14-20\\ =149-74>0\end{array}$

$\therefore S_1>0$, so P lies outside the circle joint P with centre $C(2,1)$ of the given circle.  Suppose PC cuts the circle at A and B PB is the greatest distance of  from the circle

$\begin{array}{l}PC=\sqrt{{(10-2)}^2+{(7-1)}^2}=\sqrt{8^2+6^2}=\sqrt{100}=10\\ CB=\text{radius}=\sqrt{4+1+20}=\sqrt{25}=5\\ \therefore PB=PC+CB=10+5=15\end{array}$